Question

The value of \(\int _{-π}^π\frac{2x(1+sinx)}{1+cos^2x}dx\) is...

• A. \(-\sqrt{2}π^2\)
• B. \(π^2\)
• C. \(\frac{π^2}{\sqrt{2}}\)
• D. \(-\frac{π^2}{\sqrt{2}}\)

Hint:

Split into odd and even parts; the odd part vanishes.

Answer 1

The Correct Option is B

Step 1: Split
The integrand is \(\frac{2x}{1+\cos^2x} + \frac{2x\sin x}{1+\cos^2x}\).

Step 2: Odd part
\(\frac{2x}{1+\cos^2x}\) is odd, so its integral over \([-\pi,\pi]\) is zero.

Step 3: Even part
\(\frac{2x\sin x}{1+\cos^2x}\) is even, so \(I = 4\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx\).

Step 4: Use the reflection property
For \(J = \int_0^\pi\frac{x\sin x}{1+\cos^2x}dx\), using \(x\to\pi-x\) gives \(J = \frac\pi2\int_0^\pi\frac{\sin x}{1+\cos^2x}dx\). Put \(u=\cos x\): \(\int_{-1}^1\frac{du}{1+u^2} = \frac\pi2\).

Step 5: Result
\(J = \frac{\pi^2}{4}\) and \(I = 4J = \pi^2\). Option (B).

Final Answer:
The value of the integral is pi squared. \[ \boxed{\text{(B)}\ \pi^2} \]