By using properties of definite integrals, evaluate the integrals
∫π20√sinx√sinx+√cosxdx
Let I= ∫π20√sinx√sinx+√cosxdx
Then I=∫π20√sin(π2)−xsin√(π2−x)+√cos(π2−x)dx⇒I=∫π20√cosx√cosx+√sinxdx[∵sin(π2−x)=cos x and cos(π2−x)=sin x]
On adding Eqs. (i) and (ii) , we get 2I=∫π20√sinx+√cosx√sinx+√cosxdx=∫π201dx=[x]π20=π2−0⇒1=π4