If ∫ 1/x√((1-x)/(1+x))dx=g(x)+c, g(1)=0, then g(1/2) is equal to :

Solution

Maths sir · 5 steps

Rationalise the root so the integrand splits into two standard pieces, then put x = sin θ.

1. Rationalise. Multiply top and bottom inside the root by 1-x. For 0 < x < 1 this gives √((1-x)/(1+x)) = (1-x)/√(1-x²), so the integrand becomes 1/x√(1-x²) - 1/√(1-x²).
2. Substitute. Put x = sin θ, so dx = cos θ dθ and √(1-x²) = cos θ. The first piece becomes ∫ cosecθ dθ = ln|tan θ/2|, and the second piece is sin⁻¹ x = θ.

   g(x) = ln(tan θ/2) - θ + C, x = sin θ

3. Fix the constant. At x = 1, θ = π/2 and tan π/4 = 1. So 0 - π/2 + C = 0, which gives C = π/2.
4. Put x = 1/2. Now θ = π/6 and tan π/12 = 2-√3. So g(1/2) = ln(2-√3) - π/6 + π/2 = ln(2-√3) + π/3.
5. So. Since (√3-1)/(√3+1) = (√3-1)²/2 = 2-√3, the value is ln((√3-1)/(√3+1)) + π/3, which is option (a).

The idea to keep. When you see √((1-x)/(1+x)), multiply inside by 1-x to get (1-x)/√(1-x²). The integrand then splits into standard pieces, and x = sin θ finishes the job.

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