Question

If $\int \frac{\operatorname{cosec}^2 x}{(\operatorname{cosec} x+\cot x)^{\frac{9}{2}}} d x=(\operatorname{cosec} x-\cot x)^{\frac{7}{2}}\left(\frac{1}{\alpha}+\frac{(\operatorname{cosec} x-\cot x)^2}{11}\right)+C$ 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)

$I=\int \frac{\operatorname{cosec}^2 x}{(\operatorname{cosec} x+\cot x)^{9 / 2}} d x$
Put $\operatorname{cosec} x+\cot x=t \Rightarrow \operatorname{cosec} x-\cot x=\frac{1}{t} $
$\Rightarrow \left(-\operatorname{cosec} x \cot x-\operatorname{cosec}^2 x\right) d x=d t$
$=\int \frac{\frac{1}{2}\left(t+\frac{1}{t}\right)\left(\frac{-1}{t}\right)}{t^{9 / 2}} d x=\frac{-1}{2} \int\left(\frac{-1}{t^{9 / 2}}-\frac{1}{t^{13 / 2}}\right) d x=\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 $