Q

# BITSAT The value of the integral

$\int_{0}^{\pi /2} \frac{1}{1+\sqrt{tanx}} dx$

1. $\pi$

2. $\pi$/2

3. $\pi$/4

4. 3$\pi$/2

Views

$I = \int_{0}^{\pi /2} \frac{1}{1+\sqrt{\tan x}}dx = \int_{0}^{\pi /2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$...............................(1)

We also know that, $\int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx$

Using this property in equation (1) , $I = \int_{0}^{\pi /2} \frac{\sqrt{\cos (\frac{\pi}{2} -x)}}{\sqrt{\cos (\frac{\pi}{2} -x)}+\sqrt{\sin (\frac{\pi}{2} -x)}}.dx = \int_{0}^{\pi /2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}.dx$ .....................(2)

Adding equation (1) and (2), $2I = \int_{0}^{\pi /2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}.dx + \int_{0}^{\pi /2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}.dx$

$\Rightarrow 2I = \int_{0}^{\pi /2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}.dx = \int_{0}^{\pi /2} 1.dx$

$2I = \frac{\pi}{2} - 0 \Rightarrow I = \frac{\pi}{4}$

Option (3) is correct

Exams
Articles
Questions