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):  \(\pi^2\)

Answer 2

Step 1: Split the integrand into even/odd parts
We are to evaluate \[ I=\int_{-\pi}^{\pi}\frac{2y\,(1+\sin y)}{1+\cos^2 y}\,dy = \int_{-\pi}^{\pi}\frac{2y}{1+\cos^2 y}\,dy \;+\; \int_{-\pi}^{\pi}\frac{2y\sin y}{1+\cos^2 y}\,dy. \] Note that \(y\) is an odd function, \(\cos^2 y\) is even, and \(\sin y\) is odd. Thus \[ \frac{2y}{1+\cos^2 y}\ \text{is odd} \ \Longrightarrow \ \int_{-\pi}^{\pi}\frac{2y}{1+\cos^2 y}\,dy=0, \] and the integral reduces to \[ I=\int_{-\pi}^{\pi}\frac{2y\sin y}{1+\cos^2 y}\,dy. \] Here \(\dfrac{2\sin y}{1+\cos^2 y}\) is odd, multiplied by \(y\) (odd) gives an even integrand; so symmetry does not make it vanish.

Step 2: Integration by parts
Set \(u=y\), \(du=dy\), and \(dv=\dfrac{2\sin y}{1+\cos^2 y}\,dy\). Compute \(v\): \[ v=\int \frac{2\sin y}{1+\cos^2 y}\,dy. \] Use \(t=\cos y\), \(dt=-\sin y\,dy\), then \[ v=\int \frac{2\sin y}{1+\cos^2 y}\,dy = \int \frac{-2\,dt}{1+t^2}=-2\arctan t=-2\arctan(\cos y). \] Therefore, \[ I=\Big[u\,v\Big]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}v\,du =\Big[y\cdot\big(-2\arctan(\cos y)\big)\Big]_{-\pi}^{\pi} +2\int_{-\pi}^{\pi}\arctan(\cos y)\,dy. \]

Step 3: Evaluate the boundary term
Since \(\cos(\pi)=\cos(-\pi)=-1\) and \(\arctan(-1)=-\tfrac{\pi}{4}\), \[ \Big[y\cdot\big(-2\arctan(\cos y)\big)\Big]_{-\pi}^{\pi} = \pi\cdot\Big(-2\cdot\big(-\tfrac{\pi}{4}\big)\Big) - \big(-\pi\big)\cdot\Big(-2\cdot\big(-\tfrac{\pi}{4}\big)\Big) = \frac{\pi^2}{2}-\Big(-\frac{\pi^2}{2}\Big) = \pi^2. \]

Step 4: Show the remaining integral is zero
Consider \(J=\int_{-\pi}^{\pi}\arctan(\cos y)\,dy\). The integrand is even (since \(\cos y\) is even), so \[ J=2\int_{0}^{\pi}\arctan(\cos y)\,dy. \] Use the change \(y\mapsto \pi - y\). Then \(\cos(\pi - y)=-\cos y\), and \[ \arctan(\cos(\pi - y))=\arctan(-\cos y)=-\arctan(\cos y). \] Hence the values at \(y\) and \(\pi-y\) cancel pairwise on \([0,\pi]\), giving \[ \int_{0}^{\pi}\arctan(\cos y)\,dy=0 \ \Longrightarrow \ J=0. \] Therefore, \[ I=\pi^2 + 2\cdot 0=\pi^2. \]

Final answer

\(\pi^2\)