Integration by Substitution

Hello JEE aspirants! Integration by substitution is a powerful technique to solve integrals that might seem impossible at first glance. It simplifies the integrand by replacing a part of it with a new variable, making the integration process much easier. This technique is crucial for JEE Main, as many problems are designed to test your ability to recognize and apply the correct substitution. Get ready to master this essential tool!

Conceptual Explanation and Intuition

Think of integration by substitution as the reverse of the chain rule in differentiation. The chain rule helps us find the derivative of composite functions, and u-substitution helps us find the integral of composite functions. Essentially, we're trying to "undo" the chain rule. The key idea is to identify a function and its derivative within the integral. When you see this pattern, you can use substitution to simplify the integral.

Let's say you have an integral of the form $∫f(g(x)) \cdot g'(x) dx$ . Notice that $g'(x)$ is the derivative of $g(x)$ . By substituting $u = g(x)$ , we can transform the integral into a simpler form that we can easily solve. This technique cleverly leverages the relationship between a function and its derivative to simplify complex integrals.

The Fundamental Formula

∫f(g(x)) \cdot g'(x) dx = ∫f(u) du \text{ where } u = g(x)

Let's break this down:

$f(g(x))$ : This is a composite function, where $f$ is the outer function and $g(x)$ is the inner function.
$g'(x)$ : This is the derivative of the inner function $g(x)$ .
$dx$ : Indicates that we are integrating with respect to $x$ .

Derivation/Explanation:

Let $u = g(x)$ . Then, the differential $du$ is given by:

du = g'(x) dx

Now, substitute $u$ and $du$ into the original integral:

∫f(g(x)) \cdot g'(x) dx = ∫f(u) du

This new integral is often much simpler to evaluate than the original one. Once you find the integral in terms of $u$ , remember to substitute back $g(x)$ for $u$ to get the final answer in terms of $x$ .

Example:

Consider the integral $∫2x \cdot \sin(x^2) dx$ .

Here, $g(x) = x^2$ , and $g'(x) = 2x$ . So, we can substitute $u = x^2$ and $du = 2x dx$ . The integral becomes:

∫\sin(u) du = -\cos(u) + C = -\cos(x^2) + C

Therefore, $∫2x \cdot \sin(x^2) dx = -\cos(x^2) + C$ .

Choosing the Right Substitution

Choosing the right substitution is the most crucial part of this technique. Here are some guidelines:

Look for a function and its derivative: This is the most common scenario. If you can spot a function and its derivative (or a constant multiple of its derivative), substitution is likely to work.
Inner functions of composite functions: If you have a composite function, try substituting the inner function. For example, in $∫\sin(x^3) \cdot 3x^2 dx$ , $x^3$ is a good candidate for substitution.
Complicated expressions under radicals or in denominators: If you have a complicated expression inside a square root or in the denominator of a fraction, try substituting that expression.

Example:

Consider the integral $∫\frac{x}{\sqrt{x^2 + 1}} dx$ .

Let $u = x^2 + 1$ . Then, $du = 2x dx$ , so $x dx = \frac{1}{2} du$ . The integral becomes:

∫\frac{1}{\sqrt{u}} \cdot \frac{1}{2} du = \frac{1}{2} ∫u^{-1/2} du = \frac{1}{2} \cdot 2u^{1/2} + C = \sqrt{u} + C = \sqrt{x^2 + 1} + C

Therefore, $∫\frac{x}{\sqrt{x^2 + 1}} dx = \sqrt{x^2 + 1} + C$ .

Substitution for Trigonometric Functions

Trigonometric functions often appear in integration problems, and substitution is a valuable tool for simplifying them. Common substitutions involve using trigonometric identities or recognizing derivatives of trigonometric functions.

Example:

Consider the integral $∫\sin^3(x) \cos(x) dx$ .

Let $u = \sin(x)$ . Then, $du = \cos(x) dx$ . The integral becomes:

∫u^3 du = \frac{u^4}{4} + C = \frac{\sin^4(x)}{4} + C

Therefore, $∫\sin^3(x) \cos(x) dx = \frac{\sin^4(x)}{4} + C$ .

Reverse Chain Rule

The reverse chain rule is another way to think about integration by substitution. It essentially involves recognizing that the integral is the result of applying the chain rule in reverse.

Example:

Consider the integral $∫e^{5x} dx$ .

We know that the derivative of $e^{5x}$ is $5e^{5x}$ . So, to find the integral of $e^{5x}$ , we need to divide by 5:

∫e^{5x} dx = \frac{1}{5}e^{5x} + C

This is a direct application of the reverse chain rule.

Tip: Always remember to substitute back to the original variable after integrating. Also, don't forget the constant of integration,

C

Tip: If you're stuck, try a different substitution. Sometimes the first choice isn't the best, and experimenting can lead you to the correct solution.

Common Mistakes to Avoid

Mistake: Forgetting to substitute back to the original variable. Always express your final answer in terms of the original variable.

Mistake: Incorrectly calculating the differential

du

. Double-check your derivative calculations.

Mistake: Ignoring the constant of integration,

C

. Always include

C

in indefinite integrals.

JEE-Specific Tricks

For JEE Main, time is of the essence. Here's a trick to quickly solve integrals using substitution:

Key: Look for function and its derivative

Key: Look for function and its derivative

If you can quickly identify a function and its derivative, you can often directly write the integral without explicitly performing the substitution. This comes with practice and familiarity with common derivatives.

Example:

Consider the integral $∫\frac{\sec^2(x)}{\tan(x)} dx$ .

Recognize that $\sec^2(x)$ is the derivative of $\tan(x)$ . So, the integral is simply:

∫\frac{du}{u} = \ln|u| + C = \ln|\tan(x)| + C

Therefore, $∫\frac{\sec^2(x)}{\tan(x)} dx = \ln|\tan(x)| + C$ .

By mastering integration by substitution, you'll be well-equipped to tackle a wide range of integration problems on the JEE Main. Keep practicing, and you'll become proficient at recognizing the right substitutions and applying them effectively. Good luck!