First slide

Evaluation of definite integrals

Question

$\int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ is equal to

Moderate

A

0
B

$\frac{π}{4}$
C

$\frac{π}{2}$
D

None of these

Solution

Firstly, reduce the integrand into simplest form by using
the property $\int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx$ add them and integrate.
Let
$\begin{array}{l} I = \int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx - - - (i) \\ then I = \int_{0}^{π / 2} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} dx [∵ \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ \Rightarrow I = \int_{0}^{π / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx - - - - - (ii) \\ [∵ \sin (\frac{π}{2} - x) = \cos x and \cos (\frac{π}{2} - x) = \sin x] \end{array}$
On adding Eqs. (i) and (ii), we get
$\begin{array}{l} 2 l = \int_{0}^{π / 2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \\ = \int_{0}^{π / 2} 1 dx = [x]_{0}^{π / 2} = \frac{π}{2} - 0 \\ \Rightarrow I = \frac{π}{4} \end{array}$

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