Question

Let $f\left(x\right)=\frac{1}{3}{\cot }^{3}x-\cot x+\int {\cot }^{4}xdx$ and $f(\frac{\pi }{2})=\frac{\pi }{2}$ , then $f\left(x\right)=$

• A. $\pi -x$
• B. $x-\pi$
• C. $\frac{\pi }{2}-x$
• D. $x$

Answer 1

Answer: (D)

$x$

Given $f\left(x\right)=\frac{1}{3}{\cot }^{3}x-\cot x+\int {\cot }^{4}xdx$

Now, let $I=\int {\cot }^{4}xdx=\int {\cot }^{2}x{\cot }^{2}xdx$

$=\int {\cot }^{2}x(cose{c}^{2}x-1)dx$

$=\int {\cot }^{2}xcose{c}^{2}xdx-\int {\cot }^{2}xdx$

Put $\cot x=t$

$\Rightarrow -cose{c}^{2}xdx=dt$

$I=\int -t^{2}dt-\int (cose{c}^{2}x-1)dx$

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

So, $f\left(x\right)=\frac{1}{3}{\cot }^{3}x-\cot x-\frac{{\cot }^{3}x}{3}+\cot x+x+C$

$\Rightarrow f\left(x\right)=x+C$

$f\left(\frac{\pi }{2}\right)=\frac{\pi }{2}+C$

$\Rightarrow C=0$

Hence $f\left(x\right)=x$