Question

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

• A. \( \pi^2 \)
• B. \( \frac{\pi^2}{2} \)
• C. \( \frac{\pi}{2} \)
• D. \( 2\pi^2 \)

Answer 1

The Correct Option is A

\[ \int_{-\pi}^{\pi} \frac{2y(1 + \sin y)}{1 + \cos^2 y} \, dy = \int_{-\pi}^{\pi} \frac{2y}{1 + \cos^2 y} \, dy + \int_{-\pi}^{\pi} \frac{2y \sin y}{1 + \cos^2 y} \, dy \]

The first integral is zero due to the integrand being an odd function:

\[ \int_{-\pi}^{\pi} \frac{2y}{1 + \cos^2 y} \, dy = 0 \]

The second integral is:

\[ I = \int_{-\pi}^{\pi} \frac{2y \sin y}{1 + \cos^2 y} \, dy \]

Due to symmetry, this can be simplified:

\[ I = 2 \int_{0}^{\pi} \frac{y \sin y}{1 + \cos^2 y} \, dy \]

Further manipulation leads to:

\[ I = 4 \int_{0}^{\pi} \frac{y \sin y}{1 + \cos^2 y} \, dy \]

Applying symmetry properties and integration by parts yields:

\[ I = \pi^2 \]

Therefore, the solution is Option (1): \(\pi^2\)