Question

Evaluate : `∫_0^3dx/(x + sqrt(9 - x^2))`

Answer 1

Let I = `∫_0^3dx/(x + sqrt(9 - x^2))`

Put x = 3sinθ 
∴ dx = 3cosθ dθ
When  x = 0, θ = 0 
x = 3, θ = `pi/2`

∴ I = `∫_0^(pi/2) (3cosθ)/(3sinθ + sqrt(9 - 9sin^2θ))dθ`

∴ I = `∫_0^(pi/2) (3cosθ)/(3sinθ + 3cosθ)`dθ

∴ I = `∫_0^(pi/2) cosθ/(sinθ + cosθ)dθ`

Using `∫_0^a f(x)dx = ∫_0^a f(a - x)dx`

∴ I = `∫_0^(pi/2) cos(pi/2 - θ)/(sin(pi/2 - θ) + cos(pi/2 - θ))dθ`

∴ I = `∫_0^(pi/2) sinθ/(cosθ + sinθ)dθ`  .....(ii)

Adding equations (i) and (ii) 

2I = `∫_0^(pi/2) (cosθ + sinθ)/(sinθ + cosθ)dθ`

∴ 2I = `∫_0^(pi/2) dθ`

2I = `[θ]_0^(pi/2)`

∴ 2I = `pi/2`

∴ I = `pi/4`