# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \operatorname{cosec}^{3} x d x$

Solution:

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,

$=-\operatorname{cosec} x \cot x+\int \operatorname{cosec} x\left(\operatorname{cosec}^{2} x-1\right) d x$

$=-\operatorname{cosec} x \cot x+\int \operatorname{cosec}^{3} x d x-\int \operatorname{cosec} x d x$

$I=-\operatorname{cosec} x \cot x-I+\log \left|\tan \frac{x}{2}\right|+c_{1}$

$2 I=-\operatorname{cosec} x \cot x+\log \left|\tan \frac{x}{2}\right|+c_{1}$

$I=-\frac{1}{2} \operatorname{cosec} x \operatorname{cotx}+\frac{1}{2} \log \left|\tan \frac{x}{2}\right|+c_{1}$