The value of $\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^{2} y} dy$ is

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Definite Integrals

The value of

\int_{- π}^{π} \frac{2 y ( 1 + s i n y )}{1 + c o s ^{2} y} d y

Answer: (B)

I = \int_{- π}^{π} \frac{2 y ( 1 + sin y )}{1 + cos ^{2} y} d y

. The term

\frac{2 y}{1 + cos ^{2} y}

is odd

\Rightarrow 0

. The term

\frac{2 y sin y}{1 + cos ^{2} y}

is even:

I = 2 \int_{0}^{π} \frac{2 y sin y}{1 + cos ^{2} y} d y .

Using

y \to π - y

2 I^{'} = π \int_{0}^{π} \frac{2 sin y}{1 + cos ^{2} y} d y

where

I^{'} = \int_{0}^{π} \frac{y s i n y}{1 + c o s ^{2} y} d y

; with

u = cos y

\int_{0}^{π} \frac{sin y}{1 + cos ^{2} y} d y = [tan^{- 1} u]_{- 1}^{1} ... = \frac{π}{2}

. Hence

I = 4 I^{'} = 4 \cdot \frac{π ^{2}}{4} = π^{2} .

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