Q71 / 160 · TS EAMCET 2012 · Mathematics
If $\int \sec ^2 x \operatorname{cosec}^4 x d x=-\frac{1}{3} \cot ^3 x+k \tan x$ $-2 \cot x+C$, then $k$ is equal to
- A$4$
- B$3$
- C$2$
- Correct answerD$1$
Solution
Let $I=\int \sec ^2 x \operatorname{cosec}^4 x d x$
$\begin{aligned} & =\int \frac{1}{\sin ^4 x \cos ^2 x} d x=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^4 x \cos ^2 x} d x \\ & =\int \frac{d x}{\sin ^2 x \cos ^2 x}+\int \frac{d x}{\sin ^4 x} \\ & \left.=\int \frac{\left(\sin ^2 x+\cos ^2 x\right) d x}{\sin ^2 x \cos ^2 x}+\int \operatorname{cosec}^2 x\right) \\ & =\int\left(\sec ^2 x+\operatorname{cosec}^2 x\right) d x \\ & \quad+\int \operatorname{cosec}^2 x\left(1+\sin ^2 x+\cos ^2 x\right) \\ & =\tan x-\cot x+\int \operatorname{cosec}{ }^2 x d x \\ & =\tan x-\cot x-\cot x-\frac{\cot ^3 x}{3}+C \\ & =-\frac{1}{3} \cot ^3 x+\tan x-2 \cot ^2 x+C \cot ^2 x d x\end{aligned}$
But it given that,
$\begin{aligned} & I=-\frac{1}{3} \cot ^3 x+k \tan x-2 \cot x+C \\ \therefore \quad k & =1\end{aligned}$
Hence, the correct answer is option: 'D'.