The value of the integral is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
Substitute x = cos θ, so dx = -sin θ dθ. When x=0, θ=π/2; when x=1, θ=0.
The integrand becomes √((1-cosθ)/(1+cosθ)) = √(tan²(θ/2)) = tan(θ/2).
The integral becomes ∫(π/2 to 0) tan(θ/2)·(-sinθ) dθ = ∫₀^(π/2) tan(θ/2)·sinθ dθ.
Since sinθ = 2sin(θ/2)cos(θ/2), we get tan(θ/2)·sinθ = 2sin²(θ/2) = 1 - cosθ.
So the integral = ∫₀^(π/2) (1 - cosθ) dθ = [θ - sinθ]₀^(π/2) = (π/2 - 1) - 0 = π/2 - 1.
The integrand becomes √((1-cosθ)/(1+cosθ)) = √(tan²(θ/2)) = tan(θ/2).
The integral becomes ∫(π/2 to 0) tan(θ/2)·(-sinθ) dθ = ∫₀^(π/2) tan(θ/2)·sinθ dθ.
Since sinθ = 2sin(θ/2)cos(θ/2), we get tan(θ/2)·sinθ = 2sin²(θ/2) = 1 - cosθ.
So the integral = ∫₀^(π/2) (1 - cosθ) dθ = [θ - sinθ]₀^(π/2) = (π/2 - 1) - 0 = π/2 - 1.
[AIEEE 2004]