Question

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

Answer 1

Answer: 7

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