Question

Evaluate \[ \int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{1+\cos^2x}\,dx \] is:

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

Hint:

In definite integrals over symmetric intervals, first check odd-even nature. Odd functions integrate to zero over \([-a,a]\).

Answer 1

The Correct Option is B

Step 1: Split the integral.
\[ I=\int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{1+\cos^2x}\,dx \] \[ I=\int_{-\pi}^{\pi}\frac{2x}{1+\cos^2x}\,dx + \int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2x}\,dx \]

Step 2: Use odd-even property.
The function \[ \frac{2x}{1+\cos^2x} \] is odd, so \[ \int_{-\pi}^{\pi}\frac{2x}{1+\cos^2x}\,dx=0 \] Thus, \[ I=\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2x}\,dx \] Here, \[ \frac{2x\sin x}{1+\cos^2x} \] is even, so \[ I=2\int_{0}^{\pi}\frac{2x\sin x}{1+\cos^2x}\,dx \] \[ I=4\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx \]

Step 3: Use symmetry property.
Let \[ J=\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx \] Since \[ \frac{\sin x}{1+\cos^2x} \] is symmetric about \(\frac{\pi}{2}\), we use \[ \int_0^\pi xg(x)\,dx=\frac{\pi}{2}\int_0^\pi g(x)\,dx \] So, \[ J=\frac{\pi}{2}\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx \]

Step 4: Evaluate the remaining integral.
Let \[ u=\cos x \] Then, \[ du=-\sin x\,dx \] When \[ x=0,\quad u=1 \] and when \[ x=\pi,\quad u=-1 \] Therefore, \[ \int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx = \int_{-1}^{1}\frac{du}{1+u^2} \] \[ = \left[\tan^{-1}u\right]_{-1}^{1} \] \[ =\frac{\pi}{4}-\left(-\frac{\pi}{4}\right) \] \[ =\frac{\pi}{2} \] Thus, \[ J=\frac{\pi}{2}\cdot \frac{\pi}{2} \] \[ J=\frac{\pi^2}{4} \] Now, \[ I=4J \] \[ I=4\cdot \frac{\pi^2}{4} \] \[ I=\pi^2 \]

Step 5: Final conclusion.
Hence, \[ \boxed{\pi^2} \]