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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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