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=\(∫_0^{π}{2}1.dx\)

\(⇒2I=[x]_0^\frac{π}{2}\)

\(⇒2I=\frac{π}{2}\)

\(⇒I=\frac{π}{4}\)