The value of the integral ∫₋₁¹ log ₑ(√(1 - x) + √(1 + x) )dx is equal to:

Solution

Maths sir · 4 steps

Square the bracket first: the two roots combine into 2 + 2√(1-x²), and the log splits cleanly.

1. Even integrand. Replacing x by -x only swaps √(1-x) and √(1+x), so the integrand is even. Hence I = 2∫₀¹ ln(√(1-x)+√(1+x))dx.

   ln(√(1-x)+√(1+x)) = 1/2 ln(2+2√(1-x²)) = 1/2 ln 2 + 1/2 ln(1+√(1-x²))

2. Split the integral. Multiply by 2 and integrate over [0, 1]. The constant part gives ln 2, so I = ln 2 + J, where J = ∫₀¹ ln(1+√(1-x²))dx.
3. Find J. Put x = sin θ to get J = ∫₀^(π/2) cos θ ln(1+cos θ) dθ. By parts with v = sin θ, the boundary term [sin θ ln(1+cos θ)]₀^(π/2) is 0. That leaves ∫₀^(π/2) (sin² θ)/(1+cos θ) dθ = ∫₀^(π/2) (1-cos θ) dθ = π/2 - 1.
4. So. I = ln 2 + π/2 - 1, which is option (c).

The idea to keep. When a log holds a sum of square roots, square the sum inside the log. The cross term often becomes a neat √(1-x²), and x = sin θ finishes the job.

Answer from the NTA final answer key. Solution by PrepWiser.