Question

By using the properties of definite integrals, evaluate the integral: $∫_0^\frac{π}{2}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx$

Answer 1

$∫_0^\frac{π}{2}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx$

Let I=$∫_0^\frac{π}{2}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx$…..(1)

⇒I=∫_0^\frac{π}{2}$$\frac{\sqrt{sin(\frac{π}{2-x}})}{\sqrt{sin(\frac{π}{2-x}})+\sqrt{cos(\frac{π}{2-x}})}dx $∫_0^a$ ƒ(x)dx=$∫_0^a$ƒ(a-x)}dx)

⇒I=$∫_0^\frac{π}{2}\frac{\sqrt{cosx}}{\sqrt{cos}+\sqrt{sinx}}dx$.....(2)

Addind(1)and(2),we obtain

2I=$∫_0^{π}{2}\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx$

⇒2I=$∫_0^{π}{2}1.dx$

$⇒2I=[x]_0^\frac{π}{2}$

$⇒2I=\frac{π}{2}$

$⇒I=\frac{π}{4}$