Evaluate the following integrals:

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

Adding the powers, $-3+-1=-4$

Since it is an even number, we will divide numerator and denominator by cosx

$\int \frac{1}{\sin ^{3} x \cos x} d x=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{\sin ^{2} x \cos x}{\cos ^{4} x}} d x$

$=\int \frac{\sec ^{4} x}{\tan ^{3} x} d x=\int \frac{\sec ^{2} x \sec ^{2} x}{\tan ^{3} x} d x$

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

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

Put these values in the above equation:

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

$=-\frac{t^{-2}}{2}+\log t+c\left(\right.$ 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)

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

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