JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y\) is :
- A \(\pi^2\)
- B \(\frac{\pi^2}{2}\)
- C \(\frac{\pi}{2}\)
- D \(2 \pi^2\)
Answer & Solution
Correct Answer
(A) \(\pi^2\)
Step-by-step Solution
Detailed explanation
\(\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y\) \( =\int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y+\int_{-\pi}^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y \) \( =0+2 \cdot 2 \int_0^\pi y\left(\frac{\sin y}{1+\cos ^2 y}\right) d y \)…
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