Alright class, settle down! Today, we're diving into one of the most powerful and, dare I say, elegant techniques in integral calculus: Integration by Parts. This method is your go-to friend when you encounter an integral that looks like the product of two functions – something you can't solve directly using our basic integration formulas.

Think about differentiation for a moment. If you wanted to differentiate a product of two functions, say $f(x) cdot g(x)$, you had a special rule for it, right? The Product Rule: $(uv)' = u'v + uv'$. But what about integration? If you see $int f(x) cdot g(x) dx$, can you just integrate $f(x)$ and $g(x)$ separately and multiply them? Absolutely not! That's a common misconception and a big NO!

This is exactly where integration by parts comes into play. It's essentially the "reverse product rule" for integration. Just like how the power rule helps you integrate $x^n$ and substitution helps you deal with composite functions, integration by parts gives us a systematic way to tackle integrals of products.

Let's build this concept from scratch!

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1. The Genesis: Recalling the Product Rule of Differentiation

Before we integrate, let's cast our minds back to differentiation. Remember the product rule for differentiating two functions, say $u$ and $v$, both of which are functions of $x$?

If $y = u(x) cdot v(x)$, then its derivative with respect to $x$ is given by:
$frac{d}{dx}(uv) = u frac{dv}{dx} + v frac{du}{dx}$

This rule is fundamental. We're going to use it as our starting point to derive the integration by parts formula.

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2. Deriving the Integration by Parts Formula

Now, let's take our product rule for differentiation and integrate both sides with respect to $x$:

$int frac{d}{dx}(uv) dx = int left( u frac{dv}{dx} + v frac{du}{dx}
ight) dx$

On the left-hand side, integrating a derivative simply gives us the original function (up to a constant of integration, which we'll address later):
$int frac{d}{dx}(uv) dx = uv$

On the right-hand side, we can split the integral because integration is a linear operator:
$int left( u frac{dv}{dx} + v frac{du}{dx}
ight) dx = int u frac{dv}{dx} dx + int v frac{du}{dx} dx$

Now, combining both sides, we get:
$uv = int u frac{dv}{dx} dx + int v frac{du}{dx} dx$

Let's rearrange this equation to isolate one of the integrals. Our goal is to find a way to integrate a product, so let's solve for $int u frac{dv}{dx} dx$:

$int u frac{dv}{dx} dx = uv - int v frac{du}{dx} dx$

To make this formula more concise and easier to remember, we typically use the notation $du = frac{du}{dx} dx$ and $dv = frac{dv}{dx} dx$.
So, the formula for integration by parts becomes:

The Integration by Parts Formula:

$int u , dv = uv - int v , du$

This is the golden formula! Write it down, memorize it, and understand what each part represents:

• $u$: This is a function that we choose to differentiate.


• $dv$: This is the remaining part of the integrand, which we choose to integrate.


• $du$: This is the differential of $u$ (i.e., its derivative multiplied by $dx$).


• $v$: This is the integral of $dv$.

The magic of this formula is that it transforms an integral $int u , dv$ (which we might find hard to solve) into a new integral $int v , du$ (which we hope will be simpler to solve!). If the new integral is still complex, sometimes you might even need to apply integration by parts *again*!

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3. The ILATE Rule: Choosing 'u' and 'dv' Wisely

Now, here's the crucial step that often stumps beginners: How do you choose which part of your integral should be $u$ and which should be $dv$? A good choice makes the problem much easier, while a poor choice can make it more complicated or lead you in circles.

While there's no hard-and-fast rule that works 100% of the time, there's a very helpful mnemonic called ILATE (or LIATE) that serves as a general guideline for prioritizing the choice of $u$. The function that comes first in this order should generally be chosen as $u$.

ILATE Rule (Priority for 'u'):

1. I: Inverse Trigonometric Functions (e.g., $sin^{-1}x$, $ an^{-1}x$)


2. L: Logarithmic Functions (e.g., $ln x$, $log x$)


3. A: Algebraic Functions (e.g., $x^2$, $x+1$, $sqrt{x}$)


4. T: Trigonometric Functions (e.g., $sin x$, $cos x$, $sec^2 x$)


5. E: Exponential Functions (e.g., $e^x$, $a^x$)

Why this order?

• We generally want to choose $u$ as a function that becomes simpler when differentiated. Inverse trig and logarithmic functions often simplify when differentiated.


• We also want $dv$ to be a function that is easy to integrate. Exponential and trigonometric functions are generally easy to integrate.

Important Note: This is a heuristic, a rule of thumb. It works most of the time, especially for JEE Mains level problems. For more complex JEE Advanced problems, sometimes you might need to deviate from ILATE based on experience and observation of the integral.

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4. Step-by-Step Process for Applying Integration by Parts

Let's break down the application into clear steps:

1. Identify $u$ and $dv$: Look at your integral $int f(x)g(x) dx$. Use the ILATE rule to decide which part is $u$ and which part is $dv$. Remember, $dv$ must include $dx$.


2. Find $du$: Differentiate $u$ to get $du$. This means finding $frac{du}{dx}$ and then writing $du = frac{du}{dx} dx$.


3. Find $v$: Integrate $dv$ to find $v$. When integrating $dv$, you do not need to add the constant of integration ($+C$) at this stage. We'll add it at the very end.


4. Apply the Formula: Substitute $u$, $v$, $du$, and $dv$ into the integration by parts formula: $int u , dv = uv - int v , du$.


5. Evaluate the New Integral: Solve the new integral $int v , du$. This new integral should ideally be simpler than the original one. If it's still a product, you might need to apply integration by parts again!


6. Add the Constant of Integration: Once all integrations are done, remember to add $+C$ for indefinite integrals.

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5. Let's Work Through Some Examples!

Ready for some action? Let's apply what we've learned.

Example 1: The Classic $x sin x$

Evaluate $int x sin x , dx$

Step 1: Identify $u$ and $dv$.
Our functions are $x$ (Algebraic) and $sin x$ (Trigonometric). According to ILATE, Algebraic comes before Trigonometric.
So, we choose:
$u = x$
$dv = sin x , dx$

Step 2: Find $du$.
Differentiate $u$:
$frac{du}{dx} = frac{d}{dx}(x) = 1$
So, $du = 1 , dx = dx$

Step 3: Find $v$.
Integrate $dv$:
$v = int sin x , dx = -cos x$ (Remember, no $+C$ here yet)

Step 4: Apply the Formula.
Using $int u , dv = uv - int v , du$:
$int x sin x , dx = (x)(-cos x) - int (-cos x) (dx)$
$int x sin x , dx = -x cos x - int (-cos x) dx$
$int x sin x , dx = -x cos x + int cos x , dx$

Step 5: Evaluate the New Integral.
The new integral $int cos x , dx$ is a basic integral:
$int cos x , dx = sin x$

Step 6: Combine and Add $+C$.
Substitute back into our equation:
$int x sin x , dx = -x cos x + sin x + C$

And there you have it! Notice how the complex product became a simpler one (or no product at all after the second step).

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Example 2: Another Common One, $x e^x$

Evaluate $int x e^x , dx$

Step 1: Identify $u$ and $dv$.
We have $x$ (Algebraic) and $e^x$ (Exponential). ILATE suggests Algebraic before Exponential.
So, we choose:
$u = x$
$dv = e^x , dx$

Step 2: Find $du$.
Differentiate $u$:
$du = dx$

Step 3: Find $v$.
Integrate $dv$:
$v = int e^x , dx = e^x$

Step 4: Apply the Formula.
$int x e^x , dx = (x)(e^x) - int (e^x) (dx)$
$int x e^x , dx = x e^x - int e^x , dx$

Step 5: Evaluate the New Integral.
The new integral $int e^x , dx$ is straightforward:
$int e^x , dx = e^x$

Step 6: Combine and Add $+C$.
$int x e^x , dx = x e^x - e^x + C$
We can also factor out $e^x$:
$int x e^x , dx = e^x (x - 1) + C$

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Example 3: Integrating a Lone Logarithm, $ln x$

Evaluate $int ln x , dx$

"Wait a minute, sir!" you might say, "There's no product here! It's just $ln x$." Ah, but there is! We can always multiply by 1.
So, $int ln x , dx$ can be written as $int ln x cdot 1 , dx$.

Step 1: Identify $u$ and $dv$.
We have $ln x$ (Logarithmic) and $1$ (Algebraic). ILATE suggests Logarithmic before Algebraic.
So, we choose:
$u = ln x$
$dv = 1 , dx = dx$

Step 2: Find $du$.
Differentiate $u$:
$du = frac{1}{x} , dx$

Step 3: Find $v$.
Integrate $dv$:
$v = int 1 , dx = x$

Step 4: Apply the Formula.
$int ln x , dx = (ln x)(x) - int (x) left(frac{1}{x} , dx
ight)$
$int ln x , dx = x ln x - int left(frac{x}{x}
ight) dx$
$int ln x , dx = x ln x - int 1 , dx$

Step 5: Evaluate the New Integral.
The new integral $int 1 , dx$ is simple:
$int 1 , dx = x$

Step 6: Combine and Add $+C$.
$int ln x , dx = x ln x - x + C$
This is a very important result to remember!

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6. CBSE vs. JEE Focus: Fundamentals

Aspect	CBSE Board Exams (Class 12)	JEE Main & Advanced
Emphasis	Strong focus on understanding the formula, ILATE rule, and applying it correctly for standard forms. Questions are generally direct applications.	Requires a deep conceptual understanding. While fundamentals are crucial, questions often involve multiple applications, tricky choices for 'u'/'dv', or combining with other integration techniques (substitution, partial fractions).
Complexity	Typically involves a single application of integration by parts, or sometimes two for slightly harder questions. The choice of 'u' and 'dv' is usually obvious following ILATE.	Can involve multiple applications (e.g., $int x^2 e^x dx$), cyclic integrals (e.g., $int e^x sin x dx$), or integrals requiring a 'trick' application like $int sin^{-1} x dx$ (using $1$ as $dv$). Questions can be combined with definite integrals or properties.
Derivation	The derivation of the formula from the product rule is part of the syllabus and can be asked in theory questions.	Derivation is assumed knowledge. Focus is purely on application and problem-solving.

For both CBSE and JEE, understanding the *why* behind the formula and the *how* to choose $u$ and $dv$ is absolutely paramount. Master these fundamentals, and you'll build a strong foundation for tackling more advanced problems.

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So, there you have it, the basics of Integration by Parts. It's a powerful tool, and with practice, you'll find choosing $u$ and $dv$ becomes almost intuitive. Remember, the goal is always to transform a challenging integral into a simpler one. Keep practicing, and you'll master this technique in no time!

Welcome, future engineers and mathematicians, to a deep dive into one of the most powerful techniques in Integral Calculus: Integration by Parts. Just as the product rule helps us differentiate a product of two functions, integration by parts provides a method to integrate a product of two functions when direct methods or simple substitutions fall short. It's an indispensable tool for JEE Mains & Advanced, and mastering it will unlock a vast array of integration problems.

Think of it like this: Sometimes, when you have a complex task, breaking it down into two parts – one that's easier to manage and another that gets simpler by doing something else – can make the whole process manageable. Integration by parts does exactly that with integrals.

1. The Genesis: Derivation from the Product Rule

The beauty of integration by parts lies in its direct connection to a fundamental rule of differentiation – the product rule. Let's start there.

We know that if $u(x)$ and $v(x)$ are two differentiable functions of $x$, then the product rule states:

$frac{d}{dx} [u(x)v(x)] = u(x)frac{d}{dx}v(x) + v(x)frac{d}{dx}u(x)$

For simplicity, let's write $u$ for $u(x)$ and $v$ for $v(x)$, and $u'$ for $frac{du}{dx}$, $v'$ for $frac{dv}{dx}$.

So, $frac{d}{dx} (uv) = uv' + vu'$

Now, let's integrate both sides with respect to $x$:

$int frac{d}{dx} (uv) , dx = int (uv' + vu') , dx$

The integral of a derivative brings us back to the original function (up to a constant of integration), so:

$uv = int uv' , dx + int vu' , dx$

Our goal is to find a way to integrate a product. Let's rearrange this equation to isolate one of the integrals:

$int uv' , dx = uv - int vu' , dx$

This is the essence of the formula! To make it more practical for application, we usually make a substitution:

• Let $u = u(x)$


• Let $dv = v'(x) , dx$. This implies that $v = int v'(x) , dx$.


• Then $du = u'(x) , dx$.

Substituting these into our rearranged equation, we get the celebrated Integration by Parts Formula:

$int u , dv = uv - int v , du$

This formula allows us to transform a difficult integral of a product into potentially two easier parts: one algebraic term ($uv$) and another integral ($int v , du$) which we hope is simpler than the original. The art of integration by parts lies in choosing $u$ and $dv$ wisely.

2. The Crucial Choice: What to pick for 'u' and 'dv'?

The success of integration by parts hinges entirely on making the right choice for $u$ and $dv$. A poor choice can make the new integral $int v , du$ even more complex than the original, leading you in circles!

Remember, we want two things:

1. $u$ should become simpler when differentiated to get $du$.


2. $dv$ should be easily integrable to get $v$.

To guide this choice, a popular mnemonic called LIATE (or sometimes ILATE) is used:

L - Logarithmic functions (e.g., $ln x, log x$)

I - Inverse Trigonometric functions (e.g., $arcsin x, arctan x$)

A - Algebraic functions (e.g., $x, x^2, sqrt{x}, 3x+2$)

T - Trigonometric functions (e.g., $sin x, cos x, an x$)

E - Exponential functions (e.g., $e^x, a^x$)

The general rule is: Choose $u$ as the function that comes first in the LIATE order. The remaining part of the integrand (including $dx$) is $dv$.

Why does LIATE work?

• Logarithmic and Inverse Trig functions: Their derivatives are algebraic, which are generally simpler. Also, they are often difficult to integrate directly. So, we prefer to differentiate them ($u$).


• Algebraic functions: They simplify (reduce degree) upon differentiation (e.g., $x^3 o 3x^2 o 6x o 6 o 0$). If they are part of $u$, the integral $int v , du$ becomes simpler. If they are part of $dv$, their integral becomes more complex (e.g., $x o x^2/2$).


• Trigonometric and Exponential functions: These functions generally cycle or remain similar upon repeated differentiation and integration. They are usually easy to integrate, making them good candidates for $dv$.

JEE Focus: While LIATE is a great guide, it's not a rigid rule. Sometimes, you might need to deviate based on the specific problem to achieve simplification. Practice is key to developing this intuition.

3. Step-by-Step Procedure for Integration by Parts

1. Identify the two functions in the product you need to integrate.


2. Choose $u$ and $dv$ using the LIATE rule as a guide.


3. Differentiate $u$ to find $du = u' , dx$.


4. Integrate $dv$ to find $v = int dv$. (Remember, we don't need the constant of integration here; we add it at the very end).


5. Substitute $u, v, du, dv$ into the formula: $int u , dv = uv - int v , du$.


6. Evaluate the new integral $int v , du$. This new integral should be simpler than the original. If it's not, you might have chosen $u$ and $dv$ incorrectly, or you might need to apply integration by parts again.


7. Add the constant of integration $C$ at the end.

4. Illustrative Examples

Example 1: Basic Application (Algebraic x Trigonometric)

Evaluate $int x cos x , dx$

Step-by-step:

1. Identify functions: $x$ (Algebraic) and $cos x$ (Trigonometric).


2. Choose $u$ and $dv$: According to LIATE, Algebraic comes before Trigonometric.
   
   
   
   • Let $u = x$
   
   
   • Let $dv = cos x , dx$
   
   
   
   


3. Differentiate $u$:
   
   
   
   • $du = frac{d}{dx}(x) , dx = 1 , dx = dx$
   
   
   
   


4. Integrate $dv$:
   
   
   
   • $v = int cos x , dx = sin x$
   
   
   
   


5. Apply the formula: $int u , dv = uv - int v , du$
   
   
   
   • $int x cos x , dx = x(sin x) - int (sin x) , dx$
   
   
   
   


6. Evaluate the new integral:
   
   
   
   • $int sin x , dx = -cos x$
   
   
   
   


7. Combine and add constant:
   
   
   
   • $int x cos x , dx = x sin x - (-cos x) + C$
   
   
   • $int x cos x , dx = mathbf{x sin x + cos x + C}$
   
   
   
   

Example 2: Integration of a Single Function (Logarithmic)

Evaluate $int ln x , dx$

Here, it looks like there's only one function. The trick is to consider it as a product of $ln x$ and $1$.

Step-by-step:

1. Identify functions: $ln x$ (Logarithmic) and $1$ (Algebraic).


2. Choose $u$ and $dv$: According to LIATE, Logarithmic comes before Algebraic.
   
   
   
   • Let $u = ln x$
   
   
   • Let $dv = 1 , dx = dx$
   
   
   
   


3. Differentiate $u$:
   
   
   
   • $du = frac{d}{dx}(ln x) , dx = frac{1}{x} , dx$
   
   
   
   


4. Integrate $dv$:
   
   
   
   • $v = int 1 , dx = x$
   
   
   
   


5. Apply the formula: $int u , dv = uv - int v , du$
   
   
   
   • $int ln x , dx = (ln x)(x) - int (x) left(frac{1}{x} , dx
     ight)$
   
   
   
   


6. Evaluate the new integral:
   
   
   
   • $int (x) left(frac{1}{x}
     ight) , dx = int 1 , dx = x$
   
   
   
   


7. Combine and add constant:
   
   
   
   • $int ln x , dx = x ln x - x + C$
   
   
   • $int ln x , dx = mathbf{x(ln x - 1) + C}$
   
   
   
   

Example 3: Reappearing Integral (Exponential x Trigonometric)

Evaluate $int e^x sin x , dx$

This type of integral is very common in JEE and requires applying integration by parts twice, leading to the original integral reappearing on the right side.

Step-by-step (First Application):

1. Choose $u$ and $dv$: Exponential and Trigonometric functions are last in LIATE. We can pick either for $u$, but let's stick to LIATE for now (T before E).
   
   
   
   • Let $u = sin x$
   
   
   • Let $dv = e^x , dx$
   
   
   
   


2. Differentiate $u$ and integrate $dv$:
   
   
   
   • $du = cos x , dx$
   
   
   • $v = e^x$
   
   
   
   


3. Apply the formula:
   
   
   
   • $int e^x sin x , dx = e^x sin x - int e^x cos x , dx quad ldots (1)$
   
   
   
   

Now we have a new integral $int e^x cos x , dx$, which is similar in form. We need to apply integration by parts again to this new integral.

Step-by-step (Second Application for $int e^x cos x , dx$):

1. Choose $u'$ and $dv'$ (important to be consistent with the first choice): If you picked the trig function for $u$ in the first step, pick the trig function for $u'$ again.
   
   
   
   • Let $u' = cos x$
   
   
   • Let $dv' = e^x , dx$
   
   
   
   


2. Differentiate $u'$ and integrate $dv'$:
   
   
   
   • $du' = -sin x , dx$
   
   
   • $v' = e^x$
   
   
   
   


3. Apply the formula:
   
   
   
   • $int e^x cos x , dx = e^x cos x - int e^x (-sin x) , dx$
   
   
   • $int e^x cos x , dx = e^x cos x + int e^x sin x , dx quad ldots (2)$
   
   
   
   

Notice that the original integral $int e^x sin x , dx$ has reappeared! Now substitute (2) back into (1):

Let $I = int e^x sin x , dx$

$I = e^x sin x - (e^x cos x + I)$

$I = e^x sin x - e^x cos x - I$

Now, solve for $I$ algebraically:

$2I = e^x sin x - e^x cos x$

$I = frac{e^x}{2} (sin x - cos x)$

Finally, add the constant of integration:

$int e^x sin x , dx = mathbf{frac{e^x}{2} (sin x - cos x) + C}$

JEE Focus: For repeating integrals, consistency in choosing $u$ and $dv$ in subsequent applications is paramount. If you swap roles (e.g., $u=e^x$ in the second step), you might end up with $I = I$, which is unhelpful.

5. A Special Form of Integration by Parts

There's a very important result that often simplifies integrals of a specific form. It's crucial for JEE problems.

Theorem: $int e^x [f(x) + f'(x)] , dx = e^x f(x) + C$

Derivation:

Let's consider the integral $int e^x [f(x) + f'(x)] , dx$. We can split this into two integrals:

$I = int e^x f(x) , dx + int e^x f'(x) , dx$

Now, apply integration by parts to the first integral, $int e^x f(x) , dx$.

Let $u = f(x)$ and $dv = e^x , dx$.

Then $du = f'(x) , dx$ and $v = e^x$.

So, $int e^x f(x) , dx = f(x)e^x - int e^x f'(x) , dx$.

Substitute this back into the expression for $I$:

$I = left(f(x)e^x - int e^x f'(x) , dx
ight) + int e^x f'(x) , dx$

$I = f(x)e^x - int e^x f'(x) , dx + int e^x f'(x) , dx$

The two integrals cancel each other out! So we are left with:

$I = e^x f(x) + C$

This is a powerful shortcut. Always look for this pattern when you see $e^x$ multiplied by a sum of two functions.

Example 4: Using the Special Form

Evaluate $int e^x ( an x + sec^2 x) , dx$

Here, let $f(x) = an x$. Then $f'(x) = sec^2 x$.

The integral is exactly in the form $int e^x [f(x) + f'(x)] , dx$.

Therefore, by the special form, the integral is simply:

$mathbf{e^x an x + C}$

This saves a lot of time and effort compared to applying integration by parts multiple times!

6. Integration by Parts for Definite Integrals

The integration by parts formula can also be applied to definite integrals:

$int_a^b u , dv = [uv]_a^b - int_a^b v , du$

Where $[uv]_a^b = u(b)v(b) - u(a)v(a)$. The limits of integration apply to both the $uv$ term and the new integral.

Example 5: Definite Integral Application

Evaluate $int_0^{pi/2} x sin x , dx$

From Example 1, we found that $int x sin x , dx = -x cos x + sin x$. (Note: We made $u=x, dv=sin x dx$, then $du=dx, v=-cos x$. So $uv - int v du = x(-cos x) - int (-cos x) dx = -x cos x + int cos x dx = -x cos x + sin x$).

Now, apply the limits:

$int_0^{pi/2} x sin x , dx = [-x cos x + sin x]_0^{pi/2}$

$= left( -frac{pi}{2} cosleft(frac{pi}{2}
ight) + sinleft(frac{pi}{2}
ight)
ight) - (-0 cos(0) + sin(0))$

$= left( -frac{pi}{2} (0) + 1
ight) - (0 + 0)$

$= 1 - 0 = mathbf{1}$

7. CBSE vs. JEE Focus

Aspect	CBSE/State Boards	JEE Mains & Advanced
Formula Derivation	Important for understanding and sometimes asked in theory exams.	Conceptual understanding is key; derivation less directly tested but forms the basis.
LIATE Rule	A strong guideline, generally works for most problems.	A good starting point, but sometimes requires strategic deviation or multiple applications. Understanding *why* it works is more important than just memorizing.
Problem Complexity	Direct application, single/double application, standard types like $int ln x , dx$, $int x sin x , dx$.	Can involve multiple applications, special forms ($int e^x[f(x)+f'(x)] dx$), tricky choices of $u$ and $dv$, combination with other techniques (substitution, partial fractions), and definite integrals.
Common Pitfalls	Incorrectly choosing $u$ and $dv$, errors in differentiation/integration.	Same as CBSE, plus issues with repeating integrals (consistency), handling complex algebraic expressions, and evaluating definite integral limits carefully.
Expected Skills	Accurate application of the formula.	Strategic thinking, recognizing patterns, algebraic manipulation, and efficiency in calculations.

Conclusion

Integration by Parts is a fundamental technique in calculus that equips you to handle a wide range of product integrals. Its power lies in transforming a complex integral into a simpler one. Master the LIATE rule, understand its derivation, practice different types of problems, especially those involving repeating integrals and the special $e^x[f(x)+f'(x)]$ form, and you will be well-prepared for any challenge JEE throws your way.

Welcome to the 'Mnemonics and Shortcuts' section for Integration by Parts! This topic is fundamental in Integral Calculus and frequently appears in both board exams and JEE. Mastering these quick tips will significantly boost your speed and accuracy.

1. Mnemonic for Choosing the First Function (u) in Integration by Parts

The core of Integration by Parts lies in choosing the 'first function' (u) and the 'second function' (dv). An incorrect choice can make the integral much harder, or even impossible. The famous mnemonic helps you prioritize:

• LIATE (pronounced LYE-ate) or ILATE

The function that appears earlier in the LIATE/ILATE order should be chosen as 'u' (the first function to be differentiated). The remaining function is 'dv' (the second function to be integrated).

Let's break down LIATE:

• L: Logarithmic Functions (e.g., $log x, ln x$)


• I: Inverse Trigonometric Functions (e.g., $sin^{-1}x, an^{-1}x$)


• A: Algebraic Functions (e.g., $x^n, x^2+3x, ext{polynomials}$)


• T: Trigonometric Functions (e.g., $sin x, cos x, sec^2 x$)


• E: Exponential Functions (e.g., $e^x, a^x$)

Example: For $int x sin x ,dx$:

• $x$ is an Algebraic function.


• $sin x$ is a Trigonometric function.

Since 'A' comes before 'T' in LIATE, choose $u = x$ and $dv = sin x ,dx$.

JEE & CBSE Tip: Always use LIATE/ILATE to select 'u'. It significantly simplifies the integration process.

2. Shortcut for Type: $int e^x [f(x) + f'(x)] dx$

This is a very common and critical pattern that often appears in JEE and Board exams. Recognizing it saves a lot of time.

• If an integral is of the form $int e^x [f(x) + f'(x)] dx$, its solution is simply $e^x f(x) + C$.

Example: Evaluate $int e^x (sin x + cos x) dx$

• Here, $f(x) = sin x$.


• And its derivative $f'(x) = cos x$.

So, the integral is directly $e^x sin x + C$.

JEE Tip: Always look for this pattern when $e^x$ is multiplied by a sum of two functions. It's a direct application and bypasses standard integration by parts.

3. Tabular Method (Shortcut for Repeated Integration by Parts)

When you have to apply integration by parts multiple times (e.g., $int x^3 e^x dx$ or $int x^2 cos x dx$), the tabular method is a powerful shortcut, especially for JEE.

This method works best when one function (like a polynomial) eventually differentiates to zero, and the other function can be repeatedly integrated easily (like $e^x, sin x, cos x$).

Steps for the Tabular Method:

1. Choose 'u' (the function to differentiate) using LIATE/ILATE.


2. Choose 'dv' (the function to integrate).


3. Create two columns:
   
   
   
   • Column 1 (Differentiate): List 'u' and its successive derivatives until you reach zero.
   
   
   • Column 2 (Integrate): List 'dv' and its successive integrals (matching the number of derivatives).
   
   
   
   


4. Create a third column for Alternating Signs: Start with $+$, then $-$, then $+$, and so on.


5. Multiply diagonally: The solution is the sum of the products of each entry in the 'Differentiate' column with the next entry in the 'Integrate' column, multiplied by the corresponding sign.

Example: Evaluate $int x^2 e^x dx$

Differentiate (u)	Signs	Integrate (dv)
$x^2$	$+$	$e^x$
$2x$	$-$	$e^x$
$2$	$+$	$e^x$
$0$	$-$	$e^x$

The solution is: $(x^2)(e^x) - (2x)(e^x) + (2)(e^x) + C$

Which simplifies to: $e^x(x^2 - 2x + 2) + C$.

CBSE Note: While the tabular method is not explicitly taught, understanding its underlying logic helps in solving problems. JEE Note: This is a highly effective and time-saving method for multiple applications of integration by parts.

Practice these mnemonics and shortcuts to turn tricky integration by parts problems into quick scores!

Quick Tips for Integration by Parts

Integration by Parts is a powerful technique to integrate products of functions. It's crucial for JEE Main as it frequently appears in both standalone problems and as a step within larger integral calculations. Mastery of these tips will significantly improve your speed and accuracy.

The fundamental formula for integration by parts is:
$int u , dv = uv - int v , du$

Here are some quick tips to ace integration by parts:

• 
  Choosing 'u' and 'dv' (The ILATE Rule):
  The most critical step is correctly identifying which function to choose as 'u' (the function to be differentiated) and which as 'dv' (the function to be integrated). The ILATE rule is an invaluable mnemonic for this:
  
  
  
  • Inverse Trigonometric functions ($sin^{-1}x, cos^{-1}x$, etc.)
  
  
  • Logarithmic functions ($ln x, log_a x$)
  
  
  • Algebraic functions ($x^n, P(x)$)
  
  
  • Trigonometric functions ($sin x, cos x$)
  
  
  • Exponential functions ($e^x, a^x$)
  
  
  
  Choose 'u' as the function that comes first in the ILATE sequence. The remaining part (including $dx$) becomes 'dv'. This choice often simplifies the integral $int v , du$.
  


• 
  Integrate Simple Functions First:
  Always choose 'dv' such that it's easily integrable. If 'dv' leads to a complicated integral for 'v', reconsider your choice of 'u' and 'dv'.
  


• 
  When '1' is 'dv':
  For functions like $int ln x , dx$ or $int sin^{-1}x , dx$, where there's only one function, treat '1' as the second function 'dv'.
  Example: For $int ln x , dx$, let $u = ln x$ and $dv = 1 , dx$. Then $du = frac{1}{x} , dx$ and $v = x$.
  


• 
  The $e^x(f(x) + f'(x))$ Form:
  Memorize this standard result: $int e^x [f(x) + f'(x)] , dx = e^x f(x) + C$.
  Many JEE problems are designed to fit this form after a small manipulation or substitution. Always look for this pattern when $e^x$ is present.
  


• 
  Cyclic Integrals:
  For integrals like $int e^{ax} sin(bx) , dx$ or $int e^{ax} cos(bx) , dx$, applying integration by parts twice will bring back the original integral on one side of the equation. You can then solve for the integral algebraically.
  


• 
  Polynomials with Trigonometric/Exponential Functions:
  If 'u' is a polynomial (e.g., $x^n$) and 'dv' is a trigonometric or exponential function, you might need to apply integration by parts multiple times until the polynomial term becomes a constant (derivative is zero).
  


• 
  Definite Integrals with By Parts:
  When applying integration by parts to definite integrals, remember to apply the limits to the $uv$ term immediately after integration:
  $int_a^b u , dv = [uv]_a^b - int_a^b v , du$.
  Warning: Do not forget to evaluate the $uv$ part at the limits.
  


• 
  Common Mistake: Forgetting $dx$:
  When choosing 'dv', remember it includes $dx$. So, $dv = g(x) , dx$.
  

Mastering these tips will significantly enhance your problem-solving skills in Integral Calculus, especially for JEE Main where efficiency is key. Practice frequently to make these techniques second nature.

Welcome to the "Intuitive Understanding" section for Integration by Parts! This method is a cornerstone of Integral Calculus, essential for solving integrals of products of functions. Instead of just memorizing the formula, let's understand the core idea behind it.

The Product Rule in Reverse

At its heart, Integration by Parts is simply the product rule of differentiation in reverse. Recall the product rule for differentiation:

If (u) and (v) are differentiable functions of (x), then:

[ frac{d}{dx}(uv) = u frac{dv}{dx} + v frac{du}{dx} ]

Now, let's integrate both sides with respect to (x):

[ int frac{d}{dx}(uv) ,dx = int u frac{dv}{dx} ,dx + int v frac{du}{dx} ,dx ]

The integral of a derivative simply gives back the original function (up to a constant):

[ uv = int u frac{dv}{dx} ,dx + int v frac{du}{dx} ,dx ]

Rearranging this equation to isolate one of the integrals of a product, we get:

[ int u frac{dv}{dx} ,dx = uv - int v frac{du}{dx} ,dx ]

To make it look like the standard Integration by Parts formula, we often write (dv = frac{dv}{dx} ,dx) and (du = frac{du}{dx} ,dx):

[ int u ,dv = uv - int v ,du ]

This is the famous formula for Integration by Parts. The intuition is that if you have an integral of a product (u cdot dv), you can transform it into (uv) minus a *new* integral ( int v ,du ). The goal is for this new integral to be simpler to solve than the original one.

When and Why Do We Use It?

• 
  Product of Functions: You use Integration by Parts when you encounter an integral of a product of two different types of functions (e.g., (x cdot sin x), (x^2 cdot e^x), (ln x cdot 1)).
  


• 
  Simplification: The core idea is to transform a complex integral into one that is easier to handle. We choose one function to differentiate ((u)) and another to integrate ((dv)). The success of the method hinges on making the new integral ( int v ,du ) simpler than the original ( int u ,dv ).
  

The Intuition Behind Choosing 'u' and 'dv' (The ILATE Rule)

The choice of which function to designate as (u) and which as (dv) is crucial. A common mnemonic to guide this choice is ILATE (or LIATE):

1. Inverse Trigonometric Functions ((sin^{-1} x, an^{-1} x), etc.)


2. Logarithmic Functions ((ln x, log x), etc.)


3. Algebraic Functions ((x, x^2, sqrt{x}), etc.)


4. Trigonometric Functions ((sin x, cos x), etc.)


5. Exponential Functions ((e^x, a^x), etc.)

The function that appears earlier in the ILATE list should generally be chosen as (u).

• 
  Why this order?
  
  
  
  • We want (u) to become simpler after differentiation (e.g., (ln x) becomes (frac{1}{x}), which is algebraic; (x^n) becomes (nx^{n-1})).
  
  
  • We need (dv) to be easily integrable to get (v).
  
  
  • Functions like inverse trigonometric and logarithmic functions generally simplify greatly upon differentiation, making them excellent choices for (u). Exponential and trigonometric functions often cycle or don't simplify much upon differentiation, but they are easy to integrate, making them good choices for the (dv) part.
  
  
  
  

JEE Main/CBSE Relevance: Integration by Parts is a fundamental technique for both board exams and JEE Main. Mastering the intuitive understanding and correct application of the ILATE rule is critical for solving a wide variety of integral problems.

In essence, Integration by Parts is a strategic maneuver: you trade one integral for another, hoping the new one is more tractable. It's about simplifying the integrand through differentiation and integration of its parts.

Real World Applications of Integration by Parts

Integration by parts is a powerful technique that allows us to integrate products of functions, a scenario frequently encountered in various scientific, engineering, and economic disciplines. While often presented as a mathematical tool in textbooks, its practical implications are vast, enabling the analysis of complex systems and phenomena.

Here are some key real-world applications where integration by parts proves indispensable:

• Physics and Engineering:
  
  
  
  • Work Done: Calculating the work done by a variable force acting over a distance, especially when the force function is a product of position and another varying quantity (e.g., ∫ x cos(x) dx for a spring-like force with an oscillating component).
  
  
  • Moments and Centers of Mass: Determining the moment of inertia or the center of mass for objects with non-uniform density distributions, where the density function might be a product of coordinates.
  
  
  • Electromagnetism: Solving certain electromagnetic field problems, particularly when dealing with integrals involving products of position and field strengths.
  
  
  • Signal Processing: In Fourier analysis and Laplace transforms, integration by parts is often used to simplify integrals involving products of time-domain signals and complex exponentials, crucial for analyzing and designing filters and communication systems.
  
  
  • Solving Differential Equations: It's a fundamental step in techniques like variation of parameters or finding integrating factors for certain types of differential equations that model physical systems (e.g., in circuits, mechanical oscillations).
  
  
  
  


• Probability and Statistics:
  
  
  
  • Expected Values and Moments: Calculating the expected value (E[X] = ∫ x ⋅ f(x) dx) or higher moments of continuous random variables, especially for distributions where the probability density function (PDF), f(x), makes x ⋅ f(x) suitable for integration by parts (e.g., exponential distribution, gamma distribution).
  
  
  • Moment Generating Functions: Deriving moment generating functions, which are critical for characterizing probability distributions.
  
  
  • Reliability Engineering: Analyzing the lifespan of components where failure rates are modeled by complex functions.
  
  
  
  


• Economics and Finance:
  
  
  
  • Present Value of Income Streams: Calculating the present value of a continuous income stream that varies over time, often modeled as a product of the income rate and a discount factor (e.g., ∫ P(t) ⋅ e^-rt dt, where P(t) is income and e^-rt is the discount factor).
  
  
  • Consumer and Producer Surplus: In some complex demand and supply models, integration by parts may be needed to determine these economic metrics.
  
  
  
  

CBSE vs JEE Relevance: While the specific real-world problem formulations might be beyond the direct scope of CBSE board exams, understanding these applications provides JEE aspirants with a broader perspective on the utility and power of calculus. For JEE, the focus remains on mastering the technique itself, but knowing its practical use can enhance conceptual appreciation.

In essence, whenever you encounter a scenario where a quantity needs to be summed up (integrated) and that quantity is described as a product of two functions, one of which simplifies upon differentiation and the other is easily integrable, integration by parts is your go-to technique.

Understanding complex mathematical concepts often becomes significantly easier when we can relate them to everyday situations through analogies. For 'Integration by Parts', a powerful technique for integrating products of functions, analogies can demystify its strategic application, especially the crucial choice of 'u' and 'dv'.

Analogy 1: The Layered Gift Box

Imagine you have a beautifully wrapped, but complex, gift box that you need to open. You can't just rip it open; you need a methodical approach to reveal what's inside. This situation perfectly mirrors an integral requiring Integration by Parts.

• The Complex Gift (Original Integral): Your initial integral, $int u , dv$, is like the entire, multi-layered gift box. It's too complex to open all at once using simple methods.


• Choosing Your First Move ('u' and 'dv'): Before you start, you strategize. You decide which part of the wrapping or the box's structure ('u') you will work on first to simplify it (by differentiating it). The remaining part ('dv') is what you'll initially "open" or interact with (by integrating it).
  
  
  
  • The LIATE rule (Logarithmic, Inverse Trig, Algebraic, Trigonometric, Exponential) is your guide here. It helps you choose 'u' such that when differentiated, it simplifies the overall problem, much like picking the easiest or most impactful layer of wrapping to remove first.
  
  
  
  


• Initial Unwrapping (the $uv$ term): Once you differentiate 'u' to 'du' and integrate 'dv' to 'v', the term '$uv$' represents the part of the gift you've successfully uncovered or the initial progress you've made. You've revealed something tangible.


• The Remaining, Hopefully Simpler, Gift ($int v , du$): After this initial unwrapping, you are left with a new integral, $int v , du$. The entire purpose of integration by parts is that this *new* integral should be significantly simpler to solve than the original one. It's like finding a smaller, less intricately wrapped box inside the first one. If the new integral is more complex, it's a strong indication that your initial choice for 'u' and 'dv' was not optimal, and you should re-evaluate your "unwrapping strategy."

This analogy effectively highlights the strategic decision-making involved in choosing 'u' and 'dv' and the ultimate goal of transforming a complex integral into a more manageable one.

Analogy 2: The Product Rule's Reverse

While not a real-world analogy in the same sense, it's a powerful conceptual link that directly explains the origin and nature of Integration by Parts.

• Recall the Product Rule for Differentiation: If $h(x) = f(x)g(x)$, then $h'(x) = f'(x)g(x) + f(x)g'(x)$.


• Integrating both sides with respect to $x$ gives: $int h'(x) , dx = int f'(x)g(x) , dx + int f(x)g'(x) , dx$.


• This simplifies to: $f(x)g(x) = int g(x)f'(x) , dx + int f(x)g'(x) , dx$.


• Rearranging to solve for one of the integrals: $int f(x)g'(x) , dx = f(x)g(x) - int g(x)f'(x) , dx$.


• Now, by making the substitutions:
  
  
  
  • Let $u = f(x)$ (so $du = f'(x) , dx$)
  
  
  • Let $dv = g'(x) , dx$ (so $v = g(x)$)
  
  
  
  


• We arrive at the familiar formula: $int u , dv = uv - int v , du$.

This shows that Integration by Parts is not an arbitrary formula, but rather the direct inverse operation of the product rule of differentiation, allowing us to "undo" the product rule when integrating.

Prerequisites for Integration by Parts

Before diving into the method of Integration by Parts, it is crucial to have a strong foundation in several fundamental calculus concepts. Mastering these prerequisites will ensure a smoother understanding and application of this advanced integration technique.

Here are the key concepts you should be familiar with:

• 
  1. Basic Differentiation Formulas:
  

  A thorough knowledge of standard differentiation formulas is essential. The Integration by Parts formula, $int u , dv = uv - int v , du$, requires you to differentiate the function chosen as 'u'.

  
  
  
  
  • Derivatives of algebraic functions (e.g., $x^n$, $sqrt{x}$, $1/x$).
  
  
  • Derivatives of trigonometric functions (e.g., $sin x$, $cos x$, $ an x$).
  
  
  • Derivatives of inverse trigonometric functions (e.g., $sin^{-1} x$, $ an^{-1} x$).
  
  
  • Derivatives of exponential functions (e.g., $e^x$, $a^x$).
  
  
  • Derivatives of logarithmic functions (e.g., $ln x$, $log_a x$).
  
  
  • Product Rule and Chain Rule for differentiation.
  
  
  
  


• 
  2. Basic Integration Formulas:
  

  You will need to integrate the function chosen as 'dv' to find 'v'. Therefore, proficiency in basic indefinite integration formulas is non-negotiable.

  
  
  
  
  • Integrals of algebraic functions.
  
  
  • Integrals of trigonometric functions.
  
  
  • Integrals of exponential functions.
  
  
  • Integrals leading to inverse trigonometric functions (e.g., $int frac{1}{1+x^2} dx$).
  
  
  
  


• 
  3. Substitution Method of Integration:
  

  Often, the integral $int v , du$ obtained after applying the Integration by Parts formula may itself require the method of substitution to be solved. Sometimes, even the original integral might benefit from a substitution before applying parts.

  
  

  JEE Callout: Many JEE problems combine substitution with integration by parts, so mastering both is critical.

  
  


• 
  4. Understanding of Function Types:
  

  The choice of 'u' and 'dv' in integration by parts often relies on recognizing different types of functions and their relative ease of differentiation or integration. Familiarity with the categories of functions (Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential - commonly remembered by the acronym ILATE or LIATE) is crucial for making an effective choice.

  
  


• 
  5. Algebraic Manipulation and Simplification:
  

  After applying the formula, the resulting expression may need significant algebraic manipulation to simplify or to prepare it for further integration. Strong algebraic skills are vital for avoiding errors and reaching the final solution.

  
  

By ensuring a solid grasp of these prerequisite topics, you will be well-equipped to tackle the complexities of Integration by Parts effectively, making your learning experience more productive and your problem-solving more accurate.

Mastering Integration by Parts requires a solid foundation in several other key calculus topics. These related concepts are not just prerequisites but also frequently appear hand-in-hand with integration by parts in complex problems, especially in JEE Main. Understanding these connections enhances your problem-solving ability and efficiency.

Here are the crucial related topics:

• 
  Product Rule of Differentiation:
  

  Integration by Parts is directly derived from the product rule of differentiation, which states: $(uv)' = u'v + uv'$. Rearranging and integrating this equation leads to the integration by parts formula: $int u ,dv = uv - int v ,du$. A clear understanding of the product rule provides a conceptual backbone for why integration by parts works.

  
  


• 
  Basic Integration Formulas & Techniques:
  

  To successfully apply integration by parts, you must be proficient in basic integration. This includes:

  
  
  
  
  • Standard Integrals: Knowing integrals of common functions like polynomials, exponentials ($int e^x dx$), trigonometric functions ($int sin x dx$, $int cos x dx$), and basic algebraic forms.
  
  
  • Integration by Substitution (u-substitution): Often, one of the integrals generated during integration by parts (especially $int v ,du$) might require substitution. Furthermore, some problems that initially look like they need integration by parts can be solved more simply with substitution, or vice-versa. Recognizing when to use each technique is vital.
  
  
  
  


• 
  Definite Integrals:
  

  Integration by parts is frequently applied to definite integrals. The formula for definite integration by parts is: $int_a^b u ,dv = [uv]_a^b - int_a^b v ,du$. Proficiency with evaluating definite integrals and applying limits is essential when this technique is used in combination with definite integration, which is common in JEE.

  
  


• 
  Inverse Trigonometric and Logarithmic Functions:
  

  These functions are very common candidates for 'u' when applying the ILATE/LATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) for choosing 'u' and 'dv'. Understanding their derivatives and basic integrals is crucial for setting up the integration by parts correctly.

  
  


• 
  Reduction Formulas:
  

  Many reduction formulas, which express an integral in terms of a similar integral with a lower power, are derived using integration by parts. For example, formulas for $int sin^n x ,dx$ or $int x^n e^x ,dx$ often involve repetitive application of integration by parts. This demonstrates a powerful application of the technique.

  
  

By strengthening your understanding of these related topics, you will not only be better equipped to tackle integration by parts problems but also build a more robust overall foundation in integral calculus for both CBSE board exams and the challenging JEE Main examination.

Navigating Integration by Parts in exams can be challenging, and examiners often design questions to expose common pitfalls. Being aware of these traps can significantly improve accuracy and save precious time during both CBSE board exams and JEE Main.

Here are the common exam traps related to Integration by Parts:

• 
  1. Incorrect Choice of First (u) and Second (dv) Function:
  

  The Trap: This is arguably the most common mistake. Students often choose 'u' and 'dv' arbitrarily without following the hierarchy suggested by the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential). An incorrect choice usually leads to a more complex integral in the next step, making the problem harder or even impossible to solve through this method.

  
  

  JEE Trap: Problems might be crafted where an immediate, intuitive choice (e.g., $x^2$ as 'u' when $log x$ is also present) is wrong, leading to lengthy calculations. Always prioritize ILATE.

  
  

  Tip: Remember ILATE. The function that comes earlier in ILATE should generally be chosen as 'u' (the first function) because it simplifies upon differentiation.

  
  


• 
  2. Sign Errors:
  

  The Trap: The formula is $int u , dv = uv - int v , du$. The minus sign between the $uv$ term and the new integral is frequently forgotten or mishandled, especially when 'v' or 'du' itself contains a negative sign.

  
  

  CBSE Trap: Simple sign errors can lead to incorrect final answers, costing easy marks. Double-check all signs, especially when integrating trigonometric functions or differentiating negative terms.

  
  

  Tip: Enclose the $v , du$ term in parentheses before evaluating the integral, especially if it involves multiple terms or negative signs, to prevent sign mix-ups.

  
  


• 
  3. Forgetting to Differentiate 'u' or Integrate 'dv' Correctly:
  

  The Trap: It's crucial to differentiate 'u' to get 'du' and integrate 'dv' to get 'v'. Students sometimes mix these operations up, e.g., integrating 'u' or differentiating 'dv'. Basic differentiation/integration errors are also common here.

  
  

  CBSE Trap: This often boils down to a lack of practice with fundamental differentiation/integration formulas. Ensure you are well-versed with these basics.

  
  

  Tip: Write down 'u', 'du/dx', 'dv/dx', and 'v' explicitly before substituting them into the formula. This structured approach minimizes errors.

  
  


• 
  4. Infinite Loops or Increasing Complexity:
  

  The Trap: If the choice of 'u' and 'dv' is wrong, or if integration by parts is applied mechanistically without thought, the new integral might be identical to the original one (but with opposite sign, or on the same side, creating a loop) or significantly more complex than the original. For example, applying parts to $int frac{1}{x} cdot log x , dx$ with $u=1/x$ will increase complexity.

  
  

  JEE Trap: Recognizing recurring integrals (e.g., $int e^x sin x , dx$) where the original integral reappears on the RHS is key. Students often keep applying parts indefinitely instead of treating it as an algebraic equation for the integral.

  
  

  Tip: Always assess the new integral $int v , du$. It should be simpler or solvable. If it's more complex, re-evaluate your choice of 'u' and 'dv'. For recurring integrals, set the original integral equal to 'I' and solve for 'I'.

  
  


• 
  5. Missing the '1' as a Second Function:
  

  The Trap: For integrals like $int log x , dx$ or $int an^{-1} x , dx$, students often get stuck because there seems to be only one function. They forget that '1' can always be taken as the second function ($dv$).

  
  

  CBSE/JEE Trap: This is a common oversight that stalls many students. Always consider '1' as a potential algebraic function for 'dv' when only one function type (especially inverse or logarithmic) is present.

  
  

  Tip: For $int f(x) , dx$ where $f(x)$ is a logarithmic or inverse trigonometric function, set $u = f(x)$ and $dv = 1 , dx$.

  
  


• 
  6. Incorrect Application for Special Forms:
  

  The Trap: For specific forms like $int e^x (f(x) + f'(x)) , dx = e^x f(x) + C$, students might laboriously apply integration by parts twice instead of directly using the formula. Similarly, some forms require algebraic manipulation or substitution before applying parts.

  
  

  JEE Trap: Recognising these special forms is a huge time-saver. Not identifying them means spending valuable minutes on longer methods.

  
  

  Tip: Familiarize yourself with all standard results and special integral forms to apply them judiciously.

  
  


• 
  7. Definite Integral Errors:
  

  The Trap: When applying integration by parts to definite integrals, students sometimes forget to evaluate the $uv$ term at the limits, or apply limits incorrectly only to one part of the expression.

  
  

  CBSE/JEE Trap: Remember that for $int_a^b u , dv = [uv]_a^b - int_a^b v , du$. Both terms require limit application. Don't forget the constant of integration 'C' for indefinite integrals, and remember it's not needed for definite integrals.

  
  

  Tip: Always write the limits clearly with the $uv$ term as $[uv]_a^b$ before substitution.

  
  

Mastering integration by parts comes with practice. Be mindful of these common traps and critically review your steps, especially during practice, to avoid them in the actual examination.

Welcome to the 'Problem Solving Approach' for Integration by Parts! This section will equip you with a systematic strategy to tackle problems using this powerful technique, focusing on choosing the correct functions and simplifying the process.

The Integration by Parts Formula:

Recall the fundamental formula for Integration by Parts:

$\int u dv=uv-\int v du$

The core challenge in applying this formula lies in correctly identifying which part of the integrand should be 'u' (the function to differentiate) and which part should be 'dv' (the function to integrate).

The ILATE/LIATE Rule for Choosing 'u':

To systematically choose 'u', we use a helpful mnemonic: ILATE (or sometimes LIATE). This rule prioritizes the type of function that should be chosen as 'u' based on its position in the acronym:

• Inverse Trigonometric Functions (e.g., ${\sin }^{-1}x$, tan−1x)


• Logarithmic Functions (e.g., $\ln x$, ${\log }_a x$)


• Algebraic Functions (e.g., x, x2, polynomials)


• Trigonometric Functions (e.g., sin x, cos x)


• Exponential Functions (e.g., ex, ax)

The function that appears earlier in the ILATE sequence should generally be chosen as 'u'. The remaining part of the integrand becomes 'dv'. This choice usually ensures that ∫v du is simpler to integrate than the original integral.

Systematic Problem-Solving Steps:

1. Identify the Functions: Clearly identify the two distinct functions multiplied together in the integrand.


2. Choose 'u' and 'dv': Apply the ILATE rule. The function higher in the ILATE sequence is 'u'. The remaining part (including dx) is 'dv'.


3. Calculate 'du' and 'v':
   
   
   
   • Differentiate 'u' to find 'du'.
   
   
   • Integrate 'dv' to find 'v'. (Don't include the constant of integration here; it will be added at the final step).
   
   
   
   


4. Apply the Formula: Substitute 'u', 'v', 'du', and 'dv' into the integration by parts formula: ∫u dv=uv−∫v du.


5. Evaluate the New Integral: Solve the integral ∫v du. This might require further integration techniques, including applying integration by parts again (e.g., for ∫x2ex dx).


6. Add the Constant of Integration: Once all integrals are evaluated, add the constant 'C'.

JEE Specific Tips & Special Forms:

• Integrating '1': For functions like ∫ln x dx or ∫sin−1x dx, consider '1' as the second function, i.e., ∫ln x · 1 dx. Here, u=ln x (Logarithmic), and dv=1 dx (Algebraic).


• Standard Form: ∫ex[f(x)+f(x)] dx=exf(x)+C: Always look out for this direct application. If you encounter ex multiplied by a sum of two terms, check if one is the derivative of the other. This saves significant time.


• Cyclic Integrals: For integrals like ∫eaxsin (bx) dx or ∫eaxcos (bx) dx, applying integration by parts twice will bring the original integral back. You can then solve for the integral algebraically.

CBSE vs. JEE: While the ILATE rule is fundamental for both, JEE problems often involve multiple applications of integration by parts, the recognition of the ∫ex[f(x)+f(x)] dx form, or cyclic integrals. Mastering these special cases is crucial for competitive exams.

Example:

Evaluate ∫x sin x dx.

1. Functions: x (Algebraic) and sin x (Trigonometric).


2. Choose 'u' and 'dv' (ILATE): 'A' (Algebraic) comes before 'T' (Trigonometric). So, let u=x and dv=sin x dx.


3. Calculate 'du' and 'v':
   
   
   
   • du=dx
   
   
   • v=∫sin x dx=−cos x
   
   
   
   


4. Apply the Formula:
   

   ∫x sin x dx=x(−cos x)−∫(−cos x) dx

   
   

   =−xcos x+∫cos x dx

   
   


5. Evaluate the New Integral:
   

   ∫cos x dx=sin x

   
   


6. Final Answer:
   

   ∫x sin x dx=−xcos x+sin x+C

   
   

By following these steps and keeping the ILATE rule in mind, you can systematically approach most integration by parts problems. Practice makes perfect!

CBSE Focus Areas: Integration by Parts

For CBSE board examinations, Integration by Parts is a fundamental technique in Integral Calculus that demands both conceptual understanding and meticulous execution. The questions are generally direct, focusing on the correct application of the formula and specific standard forms. Mastering these areas will ensure you score well.

1. The Core Formula

The foundation of integration by parts is the product rule of differentiation in reverse. For CBSE, it's crucial to correctly state and apply the formula:

• $int u cdot v' dx = u cdot v - int u' cdot v dx$ (where $u$ and $v$ are functions of $x$)


• Alternatively, and more commonly used in practice: $int u dv = uv - int v du$

Here, 'u' is the first function (which is differentiated) and 'dv' is the second function (which is integrated). The choice of 'u' and 'dv' is paramount for simplifying the integral.

2. The ILATE Rule for Choosing 'u' and 'dv'

The ILATE rule is a mnemonic critical for CBSE students to choose the 'u' function (first function) wisely, making the integration simpler. The preference order for 'u' is:

• I - Inverse Trigonometric Functions (e.g., $sin^{-1}x, an^{-1}x$)


• L - Logarithmic Functions (e.g., $log x, ln x$)


• A - Algebraic Functions (e.g., $x, x^2, ext{polynomials}$)


• T - Trigonometric Functions (e.g., $sin x, cos x$)


• E - Exponential Functions (e.g., $e^x, a^x$)

The function appearing earlier in the ILATE sequence should be chosen as 'u'.

3. Key Integrals and Standard Forms for CBSE

CBSE frequently tests specific forms that simplify quickly with integration by parts:

• Integrating Single Functions: Functions like $log x$, $ an^{-1}x$, $sin^{-1}x$ are integrated by taking '1' as the second function ($dv$).
  
  Example: $int log x dx = int log x cdot 1 dx$. Here, $u = log x$ (L) and $dv = 1 dx$ (A).
  


• The Special Form: Highly important for CBSE!
  
  $int e^x [f(x) + f'(x)] dx = e^x f(x) + C$
  
  Recognizing this form can save significant time. You simply need to identify $f(x)$ and its derivative $f'(x)$ within the bracket.
  


• Repeated Integration by Parts: For functions like $x^n e^{ax}$, $x^n sin(ax)$, or $e^{ax} sin(bx)$, the formula may need to be applied multiple times. For $e^{ax} sin(bx)$ or $e^{ax} cos(bx)$, the original integral reappears, allowing you to solve for it algebraically.

4. Application in Definite Integrals

When applying integration by parts to definite integrals, remember to apply the limits correctly after the integration of the $uv$ term and after the final integral is evaluated:

$int_a^b u dv = [uv]_a^b - int_a^b v du$

Careful substitution of limits for each part is crucial to avoid errors.

CBSE Exam Tips for Integration by Parts:

• Step-by-Step Clarity: Always show each step clearly, especially the choice of 'u' and 'dv' and the application of the formula.


• Sign Errors: Be extremely cautious with signs, particularly the minus sign in the formula and when dealing with derivatives/integrals of trigonometric functions.


• Practice Standard Forms: Dedicate practice to the $int e^x [f(x) + f'(x)] dx$ form and integrals of $log x$, $ an^{-1}x$, etc.


• Final Answer: Don't forget the constant of integration 'C' for indefinite integrals.

Example of a Frequently Tested CBSE Type:

Evaluate $int e^x (sin x + cos x) dx$

Solution:

This integral is of the form $int e^x [f(x) + f'(x)] dx$.

Let $f(x) = sin x$.

Then $f'(x) = cos x$.

Therefore, the integral is directly equal to $e^x f(x) + C$.

So, $int e^x (sin x + cos x) dx = e^x sin x + C$.

Keep practicing these types of problems to build confidence and speed for your CBSE board exams. You've got this!

JEE Focus Areas: Integration by Parts

Integration by Parts is a fundamental technique for integrating products of functions. For JEE Main, it's crucial not just to know the formula but also to master its strategic application, recognize common patterns, and efficiently handle specific function types.

Core Concept & Formula

The integration by parts formula is derived from the product rule of differentiation:

$int u , dv = uv - int v , du$

Here, u is a function chosen to be differentiated, and dv is a function chosen to be integrated. The key challenge lies in making the correct choice for u and dv to simplify the integral.

The LIATE Rule (for choosing 'u')

The LIATE mnemonic is a highly effective heuristic for selecting u (the first function) in the expression $int u , dv$. Choose u based on which function type appears first in the following order:

• L: Logarithmic functions (e.g., $ln x$, $log_a x$)


• I: Inverse trigonometric functions (e.g., $sin^{-1} x$, $ an^{-1} x$)


• A: Algebraic functions (e.g., $x^n$, polynomials)


• T: Trigonometric functions (e.g., $sin x$, $cos x$)


• E: Exponential functions (e.g., $e^x$, $a^x$)

JEE Tip: While LIATE is a strong guide, sometimes deviating from it might simplify the integral, especially in complex scenarios. Always aim for a simpler $int v , du$.

Key Function Types & JEE Strategies

Mastering these types is crucial for JEE:

1. Integrals of Logarithmic/Inverse Trigonometric Functions Alone:
   
   
   
   • For $int ln x , dx$ or $int sin^{-1} x , dx$, consider the second function as $1$ (an algebraic function).
   
   
   • Example: For $int ln x , dx$, take $u = ln x$ and $dv = 1 cdot dx$.
   
   
   
   


2. Repeated Integration by Parts:
   
   
   
   • Often required for integrals like $int x^n e^x , dx$, $int x^n sin x , dx$, where 'n' is a positive integer.
   
   
   • You'll apply the formula 'n' times until the polynomial term becomes a constant.
   
   
   
   


3. Cyclic Integrals:
   
   
   
   • These are integrals that return to their original form after two applications of integration by parts.
   
   
   • Common examples: $int e^{ax} sin(bx) , dx$ or $int e^{ax} cos(bx) , dx$.
   
   
   • Solve by applying integration by parts twice, then transposing the original integral to one side to solve algebraically.
   
   
   
   


4. Special JEE Forms: Recognize these immediately!




• Form 1: $int e^x [f(x) + f'(x)] , dx = e^x f(x) + C$
  
  
  
  • This is a very frequent JEE pattern. Identify $f(x)$ and its derivative $f'(x)$ within the bracket.
  
  
  
  


• Form 2: $int [f(x) + x f'(x)] , dx = x f(x) + C$
  
  
  
  • This is essentially the reverse of the product rule for differentiation $(x cdot f(x))'$.
  
  
  
  


• Form 3: $int frac{x f'(x) + f(x)}{x} dx = int f'(x) + frac{f(x)}{x} dx$ (This is less of a standard form but appears in problems involving product rule derivatives)

Common Pitfalls & How to Avoid Them

• Incorrect Selection of 'u' and 'dv': A poor choice often leads to a more complex integral, or a cyclic integral that never resolves. Always check if $int v , du$ is simpler than the original integral.


• Algebraic Errors: Be careful with signs, especially when applying the formula repeatedly.


• Forgetting the Constant of Integration: A simple yet common mistake in indefinite integrals.


• Not Recognizing Special Forms: Many JEE problems are designed to test if you can spot the $int e^x [f(x) + f'(x)] , dx$ or $int [f(x) + x f'(x)] , dx$ patterns. Look for these first!

CBSE vs. JEE Perspective

• CBSE Boards: Typically focuses on direct application of the formula for standard functions and basic cyclic integrals. The problems are generally straightforward.


• JEE Main: Expect problems that require a combination of techniques, creative choice of 'u' and 'dv', recognizing special forms, algebraic manipulation, and sometimes definite integration limits alongside integration by parts. Problems might involve substitutions before applying by parts, or multiple applications within a larger problem.

Mastering integration by parts goes beyond just memorizing the formula; it requires strategic thinking and pattern recognition. Practice a wide variety of problems, especially those involving the special JEE forms and cyclic integrals, to build confidence and speed.