Full of Math Examples
FULL OF MATH EXAMPLES
HOME LEARN EXAMPLES PROBLEMS TOOLS

Integration by substitution Examples

Here are the all examples in Integration by substitution method.

Integration by substitution is a general method for solving integration problems.

Example #1

Find the integration of sin mx using substitution method.

We know that derivative of mx is m. Thus, we make the substitution mx=t so that mdx=dt.
Therefore,

$ \quad \int \sin m x d x=\frac{1}{m} \int \sin t d t $$=-\frac{1}{m} \cos t+C $$=-\frac{1}{m} \cos m x+C $

Example #2

Find the integration of $ 2x \sin \left(x^{2}+1\right) $ using substitution method.

Here,we know that derivative of $ x^{2}+1 $ is 2$ x . $ Thus, we use the substitution $ x^{2}+1=t $ so that $ 2 x d x=d t . $ Therefore, $ \int 2 x \sin \left(x^{2}+1\right) d x$$=\int \sin t d t$$=-\cos t+C$$=-\cos \left(x^{2}+1\right)+C $

Example #3

Find the integration of $ \int x e^{x^{2}} d x $ using substitution method.

Let $ x^{2}=t$
Then, $2xdx=dt$
$ \int x e^{x^{2}} d x $$ =\frac{1}{2} \int e^{t} d t $ $ =\frac{1}{2} e^{t}+C $ $ =\frac{1}{2} e^{x^{2}}+C $

Example #4

Find the integration of $ \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} $ using substitution method.

We know that derivative of $ \tan ^{-1} x=\frac{1}{1+x^{2}} . $ So, we use the substitution $ \tan ^{-1} x=t $ so that $ \frac{d x}{1+x^{2}}=d t $ Therefore, $ \int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x$$=\int \sin t d t$$=-\cos t+C$$=-\cos \left(\tan ^{-1} x\right)+\mathrm{C} $

Example #5

Find the integration of $ \int \sin ^{3} x \cos ^{2} x d x $ using substitution method.

$\int \sin ^{3} x \cos ^{2} x d x $$=\int \sin ^{2} x \cos ^{2} x(\sin x) d x $$=\int\left(1-\cos ^{2} x\right) \cos ^{2} x(\sin x) d x $

Example #6

Find the integration of $ \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} $ using substitution method.

Derivative of $ \sqrt{x} $ is $ \frac{1}{2} x^{-\frac{1}{2}}=\frac{1}{2 \sqrt{x}} $
By using substitution $ \sqrt{x}=t $
So that $ \frac{1}{2 \sqrt{x}} d x=d t $ giving $ d x=2 t d t $

Example #7

Find $ \int \frac{\sin \sqrt{x}}{\sqrt{x}} \mathrm{d} x $ using substitution method.

Substitute $\sqrt{x}=t$
Then, $ dt =\frac{1}{2} x^{-1 / 2} \mathrm{d} x $$ =\frac{1}{2 x^{1 / 2}} \mathrm{d} x $ $ =\frac{1}{2 \sqrt{x}} \mathrm{d} x $
$ \int \frac{\sin \sqrt{x}}{\sqrt{x}} \mathrm{d} x=2 \int \sin t \mathrm{d} t $
Then ,
$ \begin{aligned} 2 \int \sin t \mathrm{d} t &=-2 \cos t+c \\ &=-2 \cos \sqrt{x}+c \end{aligned} $

Example #8

Find $ \int_{1 / 4}^{1 / 2} \frac{\cos (\pi t)}{\sin ^{2}(\pi t)} d t $ using substitution method.

Let $ u=\sin (\pi t) $ so $ d u=\pi \cos (\pi t) d t $
Then, $ d u / \pi=\cos (\pi t) d t $
We change the limits to $ \sin (\pi / 4)=\sqrt{2} / 2 $ and $ \sin (\pi / 2)=1 $
Then ,
$ \int_{1 / 4}^{1 / 2} \frac{\cos (\pi t)}{\sin ^{2}(\pi t)} d t$$=\int_{\sqrt{2} / 2}^{1} \frac{1}{\pi} \frac{1}{u^{2}} d u$$=\int_{\sqrt{2} / 2}^{1} \frac{1}{\pi} u^{-2} d u$$=\frac{1}{\pi}\left.\frac{u^{-1}}{-1}\right|_{\sqrt{2} / 2} ^{1}=-\frac{1}{\pi}+\frac{\sqrt{2}}{\pi} $

Furthur Reading