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# U Substitution

In this page, we are going to discuss about U Substitution concept. In calculus, substitution method is one of the important tool for mathematicians. Substitution is the most powerful technique for solving integration. By the use of U Substitution method, we can easily solve complicated integrals.

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## What is U Substitution?

U substitution is a technique to solve integrals or anti derivatives. It makes integration easier. We can use U substitution, if an integral contains some functions and its derivatives i.e. in the form of $\int f(x) . f'(x)dx$.
To solve such type of integration, we can use substitution method. So, it will become easier to understand. Many integrals are most easily computed by means of a change of variables.
We have $\int f(x). f'(x) dx$. To solve this,

Let $u = f(x)$. Differentiating $u$ with respect to $x$, we get

$\frac{du}{dx}$ = $f'(x)$

$du = f'(x) dx$

Then, the given integral becomes $\int f(x). f'(x) dx$ = $\int u dx$
So, $\int u dx$ = $\frac{u^{2}}{2}$ $+ C$, where $C$ is the constant of integration.

The substitution is then reversed and we get
$\int f(x). f'(x) dx$ = $\int u dx$ = $\frac{u^{2}}{2}$ $+ C$ = $\frac{(f(x))^{2}}{2}$ $+ C$

### Substitution Rule for Definite Integrals

If we have definite integral with the limits of integrations, we can solve such type of integration one of two ways,

1. To find an anti-derivative, use substitution method and after that express the result(answer) in terms of original variables and use the given limit of integration.
2. When doing the substitution, change the limits of integration. In this method, there is no need to express the answers in terms of original variables.

## U Substitution Problems

Here, we will learn about U Substitution with the help of few examples.

### U Substitution Examples

Given below are some of the examples on U Substituion.

### Solved Examples

Question 1: Solve $\int (2x + 3)^{2} dx$
Solution:
Given that $\int (2x + 3)^{2} dx$

Let $(2x + 3) = u$

Then, $2 dx = du$
$\frac{du}{2}$ = $dx$

Then, $\int $$\frac{u^{2}}{2} du = \frac{1}{2}$$\times \int u^{2} du$

= $\frac{1}{2}$$\times$$\frac{u^{3}}{3}$$+ C = \frac{1}{6}$$ \times u^{3} + C$

Replace $u$ by $(2x + 3)^{3}$
$\int (2x + 3)^{2} dx$ = $\frac{1}{6}$$[(2x + 3)^{3}] + C Question 2: Find \int (\sin x) \cos x dx Solution: Given that \int (\sin x) \cos x dx Let \sin x = u Then, \cos x dx = du So, \int u du = \frac{u^{2}}{2}$$ + C$

Replace $u$ by $\sin x$, then

$\int (\sin x) \cos x dx$ = $\frac{\sin x}{2}$$+ C Question 3: Find \int$$\frac{dx}{(2x +4)^{2}}$
Solution:
Let $(2x + 4) = u$

Then, $2 dx = du \Rightarrow dx$ = $\frac{du}{2}$

So, we have $\int $$\frac{du}{2u^{2}} = \frac{1}{2}$$\frac{-1}{u}$$+ C Replacing u by (2x + 4), we get \int$$\frac{dx}{(2x +4)^{2}}$ = $\frac{-1}{2} \frac{1}{(2x+4)}$$+ C Question 4: Evaluate \int_{0}^{1} x e^{x^{2}}dx Solution: Let x^{2} = u, then 2x dx = du When x = 0, u = 0 When x = 1, u = 1 So, we have \int_{0}^{1} x e^{x^{2}}dx = \int_{0}^{1}$$\frac{e^{u}du}{2}$

Then, $\int_{0}^{1}$$\frac{e^{u}du}{2} = \frac{1}{2}.\left [ e^{u} \right ]_{0}^{1} + C = \frac{1}{2} [e^{1} - e^{0}] + C = \frac{1}{2} [e -1] + C Question 5: Evaluate \int_{0}^{\frac{\pi }{2}}\left (\sin x \right )\cos ^{2}x dx Solution: For \int_{0}^{\frac{\pi }{2}}\left (\sin x \right )\cos ^{2}x dx, let \cos x = u Then, -\sin x dx = du When x = 0, cos 0 = 1 = u When x = \frac{\pi}{2}, then \cos$$\frac{\pi}{2}$ $= 0 = u$

So, the given integral becomes $\int_{1}^{0} u^{2}(-du)$ = $(-1) \left [ \frac{u^{3}}{3} \right ]_{1}^{0}$

= $\frac{-1}{3}$ $[0 -1]$

= $\frac{1}{3}$

## U Substitution  Practice Problems

Question 1: Evaluate $\int x^{2} \cos (x^{3} + 1) dx$
Question 2: Evaluate $\int x e^{x^{2}} dx$
Question 3: Evaluate $\int \sin ^{3}x \cos x dx$
Question 4: Evaluate $\int x {\log x} dx$
Question 5: Evaluate $\int_{2}^{3} x e^{x^{2}}dx$