Question

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

• A. 5
• B. $\frac{7}{2}$
• C. 10
• D. 7

Answer 1

Answer: (D)

$I=\int \frac{{\mathrm{cosec}}^{2}x}{(\mathrm{cosec}x+\cot x{)}^{9/2}}dx$
Put $\mathrm{cosec}x+\cot x=t\Rightarrow \mathrm{cosec}x-\cot x=\frac{1}{t}$
$\Rightarrow \left(-\mathrm{cosec}x\cot x-{\mathrm{cosec}}^{2}x\right)dx=dt$
$=\int \frac{\frac{1}{2}\left(t+\frac{1}{t}\right)\left(\frac{-1}{t}\right)}{t^{9/2}}dx=\frac{-1}{2}\int \left(\frac{-1}{t^{9/2}}-\frac{1}{t^{13/2}}\right)dx=\frac{-1}{2}\left(\frac{t}{t^{7/2}}\cdot \frac{2}{7}+\frac{2}{11\cdot t^{11/2}}\right)+C$
$\Rightarrow I=\frac{1}{2}\cdot \frac{1}{t^{7/2}}\left(\frac{2}{7}+\frac{2}{11}{\left(\frac{1}{t}\right)}^2\right)+C$
$\alpha =7$