Question

If $\int \frac{\operatorname{cosec}^{2} x}{(\text{cosec} x+\cot x)^{9 / 2}} d x $
$=(\text{cosec} x-\cot x)^{7 / 2}\left(\frac{1}{\alpha}+\frac{(\text{cosec} x-\cot x)^{2}}{11}\right)+C$
(where $C$ is constant of integration and $\alpha \in N$ ), then $\alpha$ is

Answer 1

$I=\int \frac{\text{cosec}^{2} x}{(\text{cosec} x+\cot x)^{9 / 2}} d x$
Put $\text{cosec} x+\cot x=z$
$\text{cosec} x-\cot x=\frac{1}{z}$
$-2 \text{cosec}^{2} x d x=\left(1+\frac{1}{z^{2}}\right) d z$
$\therefore I=-\frac{1}{2} \int \frac{1+\frac{1}{z^{2}}}{z^{9 / 2}} d z$
$=-\frac{1}{2}\left[\int z^{-9 / 2} d z+\int z^{\frac{-13}{2}} d z\right]$
$=-\frac{1}{2}\left[\frac{z^{-7 / 2}}{(-7)} 2+\frac{z^{-11 / 2}}{(-11)} 2\right]+C$
$=z^{\frac{-7}{2}}\left[\frac{1}{7}+\frac{z^{-3}}{11}\right]+C$
$=(\text{cosec} x-\cot x)^{\frac{7}{2}}\left(\frac{1}{7}+\frac{(\text{cosec} x-\cot x)^{2}}{11}\right)+C$