Statement-1: The value of the integral
$\int\limits_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x$ is equal to $\frac{\pi}{6}$.
Statement-2: $\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$
Answer & explanation
Correct answer: option 4
Clearly, statement-2, being a standard property, is true.
We know that $\int\limits_a^b \frac{f(x)}{f(x)+f(a+b-x)} d x=\frac{b-a}{2}$
∴ $\int\limits_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x=\int\limits_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x=\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{12}$
So, statement- 1 is not true.