Question

${\int }_0^{\frac{\pi }{2}}\frac{dx}{1+\tan x}dx$ is equal to

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

Given, $I={\int }_0^{\pi /2}\frac{dx}{1+\tan x}$
$I={\int }_0^{\pi /2}\frac{\cos x}{\sin x+\cos x}dx$ ...(i)
$I={\int }_0^{\pi /2}\frac{\cos (\pi /2-x)}{\sin (\pi -2-x)+\cos (\pi /2-x)}dx$
$={\int }_0^{\pi /2}\frac{\sin x}{\cos x+\sin x}dx$ ...(ii)
On adding Eqs. (i) and (ii), we get
$2I={\int }_0^{\pi /2}\left(\frac{\sin x+\cos x}{\sin x+\cos x}\right)$
$={\int }_0^{\pi /2}dx=\pi /2$
$\Rightarrow I=\pi /4$