Evaluate the following integrals:

Solution:

$\int \frac{1}{\sin ^{3} x \cos ^{5} x} d x=\int \sin ^{-3} x \cos ^{-5} x d x$

Adding the powers, $-3+-5=-8$

Since it is an even number, we will divide numerator and denominator by $\cos ^{8} x$

$\int \frac{1}{\sin ^{3} x \cos ^{5} x} d x=\int \frac{\frac{1}{\cos ^{8} x}}{\frac{\sin ^{2} x \cos ^{5} x}{\cos ^{8} x}} d x$

$=\int \frac{\sec ^{8} x}{\tan ^{3} x} d x=\int \frac{\sec ^{6} x \sec ^{2} x}{\tan ^{3} x} d x=\int \frac{\left(\sec ^{2} x\right)^{3} \sec ^{2} x}{\tan ^{3} x} d x$

$=\int \frac{\left(1+\tan ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{3} x} d x$

We know, $(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$

Here, $a=1$ and $b=\tan ^{2} x$

Hence, $\int \frac{\left(1+\tan ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{3} x} d x=\int \frac{\left(1+\tan ^{6} x+3 \tan ^{2} x+3 \tan ^{4} x\right) \sec ^{2} x}{\tan ^{3} x} d x$

Let $\tan x=t$, then $d t=d(\tan x)=\sec ^{2} x d x$

Put these values in above equation:

$=\int \frac{1+t^{6}+3 t^{2}+3 t^{4}}{t^{3}} d t=\int\left(t^{-3}+t^{3}+3 t^{-1}+3 t\right) d t$

$=-\frac{t^{-2}}{2}+\frac{t^{4}}{4}+3 \log t+\frac{3 t^{2}}{2}+c\left(\right.$ since $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ for any $c \neq-1$ and $\int t^{-1} d t=$ logt $)$

$=-\frac{1}{2 t^{2}}+\frac{1}{4} t^{4}+3 \log t+\frac{3}{2} t^{2}+c$

$=-\frac{1}{2 \tan ^{2} x}+\frac{1}{4} \tan ^{4} x+3 \log (\tan x)+\frac{3}{2} \tan ^{2} x+c$