Question

$\int {\text{cosec}}^{4}xdx=$

• A. $\text{cot}x+\frac{{\text{cot}}^{3}x}{3}+C$
• B. $\tan x+\frac{{\tan }^{3}x}{3}+C$
• C. $-\text{cot}x-\frac{{\text{cot}}^{3}x}{3}+C$
• D. $-\tan x-\frac{{\tan }^{3}x}{3}+C$

Answer 1

The correct option is C

$-\text{cot}x-\frac{{\text{cot}}^{3}x}{3}+C$

Explanation for Correct answer:

Finding the value for the given function:

$$\begin{gathered} \int {\text{cosec}}^{4}xdx=\int {\text{cosec}}^{2}x.{\text{cosec}}^{2}xdx \\ =\int {\text{cosec}}^{2}x.\left(1+{\text{cot}}^{2}x\right)dx\left[\because {\text{cosec}}^{2}x=1+{\text{cot}}^{2}x\right] \\ =\int {\text{cosec}}^{2}x.dx+\int {\text{cosec}}^{2}x{\text{cot}}^{2}xdx\left[\because \int {\text{cosec}}^{2}x.dx=-\mathrm{co}tx\right] \\ =-\text{cot}x-\int t^{2}dt[\text{let}t=\text{cot}x\Rightarrow dt=-\cos e{c}^{2}xdx] \\ =-\text{cot}x-\frac{t^3}{3}+C \\ =-\text{cot}x-\frac{{\text{cot}}^{3}x}{3}+C \end{gathered}$$

Hence, option (C) is the correct answer.