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# 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 #4

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}$