The value of the integral $\int\limits_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x$, is
Answer & explanation
Correct answer: option 3
Let $I=\int\limits_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x$ ...(i)
$\Rightarrow I=\int\limits_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$ ...(ii)
This is of the form $\int\limits_a^b \frac{f(x)}{f(x)+f(a+b-x)} d x$
∴ $I=\frac{\pi / 3-\pi / 6}{2}=\frac{\pi}{12}$