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Find the valu...

Question

Find the value of $I = \int_{0}^{π / 2} \frac{\sqrt{sin x}}{\sqrt{sin x} + \sqrt{cos x}} d x$

A

\frac{π}{4}

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B

\frac{π}{3}

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C

\frac{π}{6}

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D

None of these

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Solution

The correct option is B $\frac{π}{4}$
$I = \int_{0}^{π / 2} \frac{\sqrt{sin x}}{\sqrt{sin x} + \sqrt{cos x}} d x = \int_{0}^{π / 2} \frac{\sqrt{sin (\frac{π}{2} - x)}}{\sqrt{sin (\frac{π}{2} - x)} + \sqrt{cos (\frac{π}{2} - x)}} d x$ ....(1)

Using $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

$\Rightarrow I = \int_{0}^{π / 2} \frac{\sqrt{cos x}}{\sqrt{cos x} + \sqrt{sin x}} d x$ ...(2)

Adding (1) and (2)
$2 I = \int_{0}^{π / 2} (\frac{\sqrt{sin x}}{\sqrt{sin x} + \sqrt{cos x}} + \frac{\sqrt{cos x}}{\sqrt{cos x} + \sqrt{sin x}}) d x$

$= \int_{0}^{π / 2} (\frac{\sqrt{sin x} + \sqrt{cos x}}{\sqrt{sin x} + \sqrt{cos x}}) d x = \int_{0}^{π / 2} d x = {[x]}_{0}^{π / 2} = \frac{π}{2}$

$∴ I = \frac{π}{4}$

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