If ∫ eˣ((x sin⁻¹ x)/√(1-x²)+(sin⁻¹ x)/(1-x²)^(3/2)+x/(1-x²))dx=g(x)+C, where C is the constant of integration, then g(1/2) equals :

Solution

Maths sir · 4 steps

Spot eˣ(f + f'): take the first term as f, and the other two are exactly its derivative.

1. The standard form. If the integrand is eˣ(f(x) + f'(x)), the integral is eˣ f(x) + C. So try f(x) = (x sin⁻¹ x)/√(1-x²), the first term in the bracket.
2. Differentiate f. By the product rule, d/dx[x/√(1-x²)] = 1/√(1-x²) + x²/(1-x²)^(3/2) = 1/(1-x²)^(3/2). The other part gives x/√(1-x²) × 1/√(1-x²) = x/(1-x²).

   f'(x) = (sin⁻¹ x)/(1-x²)^(3/2) + x/(1-x²)

3. Read off g. This f'(x) is exactly the second and third terms in the bracket, so g(x) = (eˣ x sin⁻¹ x)/√(1-x²).
4. So. At x = 1/2, sin⁻¹ 1/2 = π/6 and √(1-1/4) = √3/2. So g(1/2) = √e × (1/2 × π/6)/(√3/2) = π/6√(e/3), which is option (a).

The idea to keep. When you see eˣ times a messy bracket, do not integrate term by term. Pick one term as f, differentiate it, and check whether the remaining terms add up to f'.

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