Question

If $\int {\cot }^{5}xdx=A{\cot }^{4}x+B{\cot }^{2}x+\log f\left(x\right)+c$, then $[A,B,f(x\left)\right]$ is equal to

• A. $\frac{1}{4},\frac{1}{2},\sin x$
• B. $-\frac{1}{4},\frac{1}{2},-\sin x$
• C. $-\frac{1}{4},\frac{1}{2},\sin x$
• D. $\frac{1}{3},\frac{1}{4},-\sin x$

Answer 1

Answer: (C)

$-\frac{1}{4},\frac{1}{2},\sin x$

$\int {\cot }^{5}xdx=\int {\cot }^{3}x{\cot }^{2}xdx$
$=\int {\cot }^{3}x[{\text{cosec}}^{2}x-1]dx$
$=\int {\cot }^{3}x{\text{cosec}}^{2}x-\int {\cot }^{3}xdx$
$=-\int {\cot }^{3}xd\left(\cot x\right)-\int {\cot }^{3}xdx$
$=\int \frac{-{\cot }^{4}x}{4}-\int {\cot }^{2}x\cot xdx$
$=-\frac{{\cot }^{4}x}{4}-\int \cot x({\text{cosec}}^{2}x-1)dx$
$=-\frac{{\cot }^{4}x}{4}+\int \cot xd\left(\cot x\right)+\int \cot xdx$
$=\frac{-{\cot }^{4}x}{4}+\frac{{\cot }^{2}x}{2}+\log \left|\sin x\right|+c$

So, $A=-\frac{1}{4};B=\frac{1}{2};f\left(x\right)=\sin x$