Question

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

• A. $\pi -x$
• B. $x-\pi$
• C. $\frac{\pi }{2}-x$
• D. none of these

Answer 1

Answer: (D)

none of these

$\varphi \left(x\right)=\int {\cot }^{4}xdx+\frac{1}{3}{\cot }^{3}x-\cot x$
Now, $\int {\cot }^{4}xdx$
$=\int {\cot }^{2}x({\csc }^{2}x-1)dx$
$=\int {\cot }^{2}x{\csc }^{2}xdx-\int {\cot }^{2}xdx$
Put $\cot x=t$
$\Rightarrow -{\csc }^{2}xdx=dt$
$=\int -t^{2}dt-\int ({\csc }^{2}x-1)dx$
$=-\frac{t^3}{3}+\cot x+x+C$
$=-\frac{{\cot }^{3}x}{3}+\cot x+x+C$
$\varphi \left(x\right)=-\frac{{\cot }^{3}x}{3}+\cot x+x+C+\frac{1}{3}{\cot }^{3}x-\cot x$
Hence, $\varphi \left(x\right)=x+C$
$\varphi \left(\frac{\pi }{2}\right)=\frac{\pi }{2}+C$
$\Rightarrow C=0$ ($\because \varphi \left(\frac{\pi }{2}\right)=\frac{\pi }{2}$given)
Hence, $\varphi \left(x\right)=x$