Question

${\int }_0^{\frac{\pi }{4}}{\left(\frac{x}{x\sin x+\cos x}\right)}^{2}dx=$

• A. $\frac{5-\pi }{5-\pi }$
• B. $\frac{2}{5+\pi }$
• C. $\frac{4-\pi }{4+\pi }$
• D. $\frac{4+\pi }{4-\pi }$

Answer 1

Answer: (C)

$\frac{4-\pi }{4+\pi }$

${\int }_0^{\frac{\pi }{4}}x\text{sec}x\frac{x\cos x}{(x\sin x+\cos x{)}^2}dx$

${\left[x\text{sec}x\frac{-1}{(x\sin x+\cos x{)}^2}\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}(\text{sec}x+x\text{sec}x\tan x)\frac{-1}{x\sin x+\cos x}dx$

${\left[x\text{sec}x\frac{-1}{(x\sin x+\cos x{)}^2}\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}{\text{sec}}^{2}xdx$

${\left[x\text{sec}x\frac{-1}{(x\sin x+\cos x{)}^2}\right]}_0^{\frac{\pi }{4}}+{\left[\tan x\right]}_0^{\frac{\pi }{4}}$

${[\frac{-x}{\cos x{(x\sin x+\cos x)}^2}+\frac{\sin x}{\cos x}]}_0^{\frac{\pi }{4}}$

$⟹{\left[\frac{\sin x-x\cos x}{x\sin x+\cos x}\right]}_0^{\frac{\pi }{4}}$

On Substituting limits

$\frac{\frac{1}{\sqrt{2}}-\frac{\pi }{2}\left(\frac{1}{\sqrt{2}}\right)}{\frac{\pi }{4}\left(\frac{1}{\sqrt{2}}\right)+\frac{1}{\sqrt{2}}}$

$⟹\frac{4-\pi }{4+\pi }$