Question

$\int_\limits{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ is equal to

• A. $0$
• B. $ - \pi $
• C. $3 \pi / 2$
• D. $ \pi / 2$
• E. $\pi / 4$

Answer 1

Answer: (E)

Let $I=\int\limits_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos X}} d x\,\,\,...(i)$
$\Rightarrow I=\int\limits_{0}^{\pi / 2} \frac{\sqrt{\sin (\pi / 2-x)}}{\sqrt{\sin (\pi / 2-x)}+\sqrt{\cos (\pi / 2-x)}} d x$
$\left[\because \int\limits_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(a +b -x) d x\right]$
$\Rightarrow I=\int\limits_{0}^{\pi / 2} \frac{\sqrt{{\cos}\, x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \,\,\,...(ii)$
On adding Eqs. (i) and (ii), we get
$2 I=\int\limits_{0}^{\pi / 2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
$\Rightarrow 2 \,I=\int\limits_{0}^{\pi / 2} 1 \,d x$
$\Rightarrow 2 \,I=[x]_{0}^{\pi / 2}$
$\Rightarrow 2\, I=\frac{\pi}{2} $
$\Rightarrow I=\frac{\pi}{4}$