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

Answer: (A)

$\pi^2$

Answer 2

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

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): $\pi^2$