Question

The value of ${\int }_0^{\pi /2}\frac{x+\sin x}{1+\cos x}dx$ is

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

Answer 1

The correct option is C $\frac{\pi }{2}$

Let $I={\int }_0^{\frac{\pi }{2}}\frac{x+sinx}{1+cosx}dx$
Substituting $tan\left(\frac{x}{2}\right)=t\Rightarrow \frac{1}{2}{sec}^2\left(\frac{x}{2}\right)dx=dt$
We get $sinx=\frac{2t}{1+t^2},cosx=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$
Therefore
$I={\int }_0^1\frac{2{tan}^{-1}t+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}={\int }_0^1(2{tan}^{-1}t+\frac{2t}{1+t^2})dt={\int }_0^{1}2{tan}^{-1}tdt+{\int }_0^1\frac{2t}{1+t^2}dt=2{[t{tan}^{-1}t-\frac{1}{2}\log (t^2+1)]}_0^1+{\left[\log (t^2+1)\right]}_0^1=2(\frac{\pi }{4}-\frac{1}{2}\log 2-0+0)+(\log 2+0)=\frac{\pi }{2}$