Let the solution curve y=f(x) of the differential equation (d y)/(d x)+(x y)/(x²-1)=(x⁴+2 x)/√(1-x²), x∈(-1,1) pass through the origin. Then ∫_(-√3/2)^(√3/2) f(x) d x is equal to

Solution

Maths sir · 5 steps

Integrating factor √(1-x²) gives f, and then the odd part vanishes on the symmetric limits.

1. Integrating factor. Here P = x/(x²-1), so ∫ P dx = 1/2 ln|x²-1| = 1/2 ln(1-x²) on (-1,1). The integrating factor is e^(∫ P dx) = √(1-x²).

   y√(1-x²) = ∫ (x⁴+2x)/√(1-x²)·√(1-x²) dx = x⁵/5 + x² + C

2. Use the origin. Putting x=0, y=0 gives C = 0. So f(x) = x⁵/5√(1-x²) + x²/√(1-x²).
3. Drop the odd part. The first term is odd, so its integral from -√3/2 to √3/2 is 0. The second term is even, so the answer is 2∫₀^(√3/2) x²/√(1-x²) dx.
4. Trig substitution. Put x = sin θ. The limits become 0 to π/3 and the integrand becomes sin² θ. Then ∫₀^(π/3)sin² θ dθ = [θ/2 - (sin 2θ)/4]₀^(π/3) = π/6 - √3/8.
5. So. The integral is 2(π/6 - √3/8) = π/3 - √3/4, which is option (b).

The idea to keep. Before integrating over limits like -a to a, split the function into odd and even parts. The odd part gives zero, and the even part becomes twice the integral from 0 to a.

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