Question

The value of ${\int }_{\frac{\pi }{5}}^{\frac{3\pi }{10}}\frac{\cos x}{\sin x+\cos x}dx$ equals

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. None of these

Answer 1

The correct option is D None of these

Let $I={\int }_{\frac{\pi }{5}}^{\frac{3\pi }{10}}\frac{\cos x}{\cos x+\sin x}dx$
Multiply numerator and denominator by $se{c}^3\left(x\right)$
$I={\int }_{\frac{\pi }{5}}^{\frac{3\pi }{10}}\frac{se{c}^{2}x}{se{c}^{2}x+se{c}^{2}x\tan x}dx={\int }_{\frac{\pi }{5}}^{\frac{3\pi }{10}}\frac{se{c}^{2}x}{1+\tan x+{\tan }^{2}x+{\tan }^{3}x}dx$
Put $t=\tan x\Rightarrow dt=se{c}^{2}xdx$
$\therefore I={\int }_{\tan \frac{\pi }{5}}^{\tan \frac{3\pi }{10}}\frac{1}{u^2+u^3+u+1}du={\int }_{\tan \frac{\pi }{5}}^{\frac{3\pi }{10}}(\frac{1-u}{2(u^2+1)}+\frac{1}{2(u+1)})du$
$=\frac{1}{2}{\int }_{\tan \frac{\pi }{5}}^{\tan \frac{3\pi }{10}}(\frac{1}{u^2+1}-\frac{u}{u^2+1})du+\frac{1}{2}{\int }_{\tan \frac{\pi }{5}}^{\tan \frac{3\pi }{10}}u+1du$
$={[-\frac{1}{4}\log (u^2+1)+\frac{1}{2}\log (u+1)+\frac{1}{2}{\tan }^{-1}u]}_{\log \frac{\pi }{5}}^{\log \frac{3\pi }{10}}=\frac{\pi }{20}$