# Integration By Substitution

## Integration By Substitution

Here we discuss on of the common method of Integration which is called Integration By Substitution
Let
$I = \int f(x) \; dx$
Let us substitute
x= g(t)
then
dx= g'(t) dt
Therefore
$I = \int f(g(t)) g'(t) \; dt$
Lets see few usage of this

A. if $\int f(x) \; dx = g(x)$ then $\int f(ax+ b) \; dx= \frac {1}{a} g(x)$
Proof
let us substitute
ax+b =t
then
a dx=dt
or
$dx=\frac {dt}{a}$
Therefore
$\int f(ax+ b) \; dx= \frac {1}{a} \int f(t) \; dt = \frac {1}{a} g(t)= \frac {1}{a} g(t)=\frac {1}{a} g(ax+b)$

Formulas Based on this
$\int (ax+b)^n \; dx= \frac {1}{a} \frac {(ax+ b)^{n+1}}{n+1} + C$
$\int e^{ax+b} \; dx=\frac {1}{a} e^{ax+b} + C$
$\int (\frac {1}{ax+b}) \; dx= \frac {1}{a} ln |ax +b| + c$
$\int a^{bx+c} \; dx= \frac {1}{b} \frac {a^{bx+c}}{ log a} + C$
$\int \cos (ax+b) \; dx= \frac {1}{a} \sin (ax+b) + C$
$\int \sin (ax+b) \; dx= - \frac {1}{a} \cos (ax+b) + C$
$\int \sec^2 (ax+b) \; dx= \frac {1}{a} \tan (ax +b) + C$
$\int \csc^2 (ax+b) \; dx= - \frac {1}{a} \cot^2 (ax+b)+ C$
$\int \tan (ax+b) \; dx=- \frac {1}{a} ln |\cos (ax+b)| + C$
$\int \cot (ax+b) \; dx= \frac {1}{a} ln |\sin (ax+b)| + C$
$\int \sec (ax+b)\; dx =\frac {1}{a} ln |\sec (ax+b) + \tan (ax+b)| + C$
$\int \csc (ax+b) \; dx= \frac {1}{a} ln |\csc (ax+b) - \cot (ax+b)| + C$

B.  $\int \frac {f^{'} (x)}{f(x)} \; dx = ln | f(x)| + C$
Proof
let us substitute
f(x) =t
then
f'(x) dx=dt
Therefore
$\int \frac {f^{'} (x)}{f(x)}\; dx = \int \frac {1}{t} \; dt =ln |t| + C= ln | f(x)| + C$

Example
$\int \frac {1}{1 + e^{-x}} \; dx = \int \frac {1}{1 + 1/e^x} dx = \int \frac {e^x}{1+ e^x} \; dx$
Now this above form
$= ln |1 + e^x| + C$

C.  $\int [f(x)]^n f^{'} x \; dx = \frac { [f(x)]^{n+1}}{n +1 } + C , n \ne -1$
Proof
let us substitute
f(x) =t
then
f'(x) dx=dt
Therefore
$\int [f(x)]^n f^{'} x \; dx = \int t^n \; dt =\frac { [t]^{n+1}}{n +1 }= \frac { [f(x)]^{n+1}}{n +1 } + C$

## Solved Examples

Example 1
$\int 2x \cos(x^2) \, dx$
Solution
$u = x^2$ $\frac{du}{dx} = 2x$ $dx = \frac{du}{2x}$ Now, substitute into the integral:
$\int \cos(u) \, \frac{du}{2}$ $= \frac{1}{2} \int \cos(u) \, du$ $= \frac{1}{2} \sin(u) + C$ Therefore integral is equal to
$= \frac{1}{2} \sin(x^2) + C$
Example 2
$\int \frac{\ln(x)}{x} \, dx$
Solution
$u = \ln(x)$ $\frac{du}{dx} = \frac{1}{x}$ $dx = x \, du$ Substitute in:
$\int u \, x \, du$ $= \int u \, du$ $= \frac{u^2}{2} + C$ Therefore integral is equal to
$= \frac{(\ln(x))^2}{2} + C$
Example 3
$\int \sqrt{1 - x^2} \, dx$
Solution
Here, we can use the trigonometric substitution:
$x = \sin(\theta)$ $dx = \cos(\theta) \, d\theta$ Therefore
$\int \sqrt{1 - \sin^2(\theta)} \cos(\theta) \, d\theta$ $= \int \cos^2(\theta) \, d\theta$ This can be integrated using the half-angle formula:
$\int \frac{1 + \cos(2\theta)}{2} \, d\theta$ $= \frac{\theta}{2} + \frac{\sin(2\theta)}{4} + C$ Now, reverting the substitution:
$\sin^{-1}(x) = \theta$ $2\sin(\theta)\cos(\theta) = \sqrt{1-x^2}$ Therefore
$= \frac{\sin^{-1}(x)}{2} + \frac{x \sqrt{1 - x^2}}{2} + C$

Example 4
$\int \frac{x}{\sqrt{x^2 + 4}} \, dx$
Solution
$u = x^2 + 4$ $\frac{du}{dx} = 2x$ $\frac{du}{2} = x \, dx$ Substituting into the integral:
$\int \frac{1}{\sqrt{u}} \, \frac{du}{2}$ $= \frac{1}{2} \int u^{-1/2} \, du$ $= \frac{1}{2} \cdot \frac{u^{1/2}}{1/2} + C$ $= \sqrt{u} + C$ Therefore
$= \sqrt{x^2 + 4} + C$

Example 5
$\int e^{3x} \, dx$
Solution
$u = 3x$ $du = 3 \, dx$ $\frac{du}{3} = dx$ Substituting:
$\int e^u \, \frac{du}{3}$ $= \frac{1}{3} \int e^u \, du$ $= \frac{1}{3} e^u + C$ Therefore
$= \frac{1}{3} e^{3x} + C$

Example 6
$\int \sin^2(x) \cos(x) \, dx$
Solution
$u = \sin(x)$ $du = \cos(x) \, dx$ Substituting:
$\int u^2 \, du$ $= \frac{1}{3} u^3 + C$ Therefore
$= \frac{1}{3} \sin^3(x) + C$

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