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Army Group X Practice Set 2

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Question : 59 of 70

Marks: +1, -0

\int\limits_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} \, dx

is equal to

$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\pi$
$2\pi$

Solution:

(a) Let

I=\int\limits_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} \, dx \qquad \ldots \quad \text{(i)}

=\int\limits_{0}^{\pi/2} \frac{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}+\sqrt{\cos\left(\frac{\pi}{2}-x\right)}} \, dx

\left[\because \int\limits_{0}^{a} f(x) \, dx = \int\limits_{0}^{a} f(a-x) \, dx\right]

I=\int\limits_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} \, dx

.....(ii)

On adding Eq. (i) and Eq. (ii),

we get

2I = \int\limits_{0}^{\pi/2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} \, dx

= \int\limits_{0}^{\pi/2} 1 \, dx

\Rightarrow 2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} - 0

\Rightarrow I = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}

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