Evaluate the following integrals:
$\int \operatorname{cosec}^{3} x d x$
Let $I=\int \operatorname{cosec}^{3} x d x$
$=\int \operatorname{cosec} x \times \operatorname{cosec}^{2} x d x$
Using integration by parts,
$=\operatorname{cosec} x \int \operatorname{cosec}^{2} x d x-\int \frac{d}{d x} \operatorname{cosec} x \int \operatorname{cosec}^{2} x d x$
We know that, $\int \operatorname{cosec}^{2} x d x=-\cot x$ and $\frac{d}{d x} \operatorname{cosec} x=\operatorname{cosec} x \cot x$
$=\operatorname{cosec} x x-\cot x+\int \operatorname{cosec} x \cot x-\cot x d x$
$=-\operatorname{cosec} x \cot x+\int \operatorname{cosec} x \cot ^{2} x d x$
Using integration by parts,
$=-\operatorname{cosec} x \cot x+\int \operatorname{cosec} x\left(\operatorname{cosec}^{2} x-1\right) d x$
$=-\operatorname{cosec} x \cot x+\int \operatorname{cosec}^{3} x d x-\int \operatorname{cosec} x d x$
$I=-\operatorname{cosec} x \cot x-I+\log \left|\tan \frac{x}{2}\right|+c_{1}$
$2 I=-\operatorname{cosec} x \cot x+\log \left|\tan \frac{x}{2}\right|+c_{1}$
$I=-\frac{1}{2} \operatorname{cosec} x \operatorname{cotx}+\frac{1}{2} \log \left|\tan \frac{x}{2}\right|+c_{1}$