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} $