If the value of the integral
∫₀^(1/2) x²/( 1 - x² )^(3/2) dx
is k/6, then k is equal to :

Solution

Maths sir · 4 steps

See 1 - x² under a power? Put x = sin θ and the whole integrand collapses to tan² θ.

1. Substitute. Let x = sin θ, so dx = cos θ dθ and 1 - x² = cos² θ. The limits change too: x = 0 gives θ = 0, and x = 1/2 gives θ = π/6.

   ∫₀^(π/6) (sin² θ)/(cos³ θ) cos θ dθ = ∫₀^(π/6) tan² θ dθ = ∫₀^(π/6) (sec² θ - 1) dθ

2. Integrate. The antiderivative is tan θ - θ. At the limits this gives tan π/6 - π/6 = 1/√3 - π/6.
3. Match the form. Write 1/√3 = √3/3 = 2√3/6. So the integral is (2√3 - π)/6.
4. So. Comparing with k/6 gives k = 2√3 - π, which is option (d).

The idea to keep. For √(1 - x²) or (1 - x²)^(n/2), put x = sin θ and change the limits at once. Then use tan² θ = sec² θ - 1 to integrate.

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