The value of ∫-π^π (2 y(1+sin y))/(1+cos ² y) d y is :

Solution

Maths sir · 4 steps

Split into odd and even parts first. The odd part dies, and the even part falls to the a+b-x trick.

1. Split the integrand. Write it as 2y/(1+cos² y) + (2y sin y)/(1+cos² y). The first part is odd, because y is odd and cos² y is even. The second part is even, because y sin y is even.
2. Use symmetry. On [-π, π] the odd part integrates to 0. The even part gives twice its integral on [0, π], so the value is 4I, where I = ∫₀^π (y sin y)/(1+cos² y) dy.
3. Reflect the variable. Put y → π - y. Since sin(π - y) = sin y and cos²(π - y) = cos² y, you get I = ∫₀^π ((π - y)sin y)/(1+cos² y) dy. Add this to the original I.

   2I = π∫₀^π (sin y)/(1+cos² y) dy = π[-tan⁻¹(cos y)]₀^π = π(π/4+π/4) = π²/2

4. So. I = π²/4, and the integral is 4I = π², which is option (b).

The idea to keep. On symmetric limits, always split into odd and even parts first. If an x sits next to a function of sin x, use x → a+b-x so the x cancels and leaves a constant.

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