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 represents an odd function, so:

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

Now consider the second integral:

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

We can rewrite this as:

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

Using the symmetry properties and integrating by parts, we find:

\[ I = \pi^2 \]

Thus, the answer is Option (1):  \(\pi^2\)