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: Even and odd functions
Over a symmetric interval, odd terms integrate to zero. Only $\frac{2x\sin x}{1+\cos^2x}$ survives, and it is even.

Step 2: Evaluate
$I = 2\int_0^\pi\frac{2x\sin x}{1+\cos^2x}dx = 4\cdot\frac\pi2\cdot\frac\pi2 = \pi^2$.

Step 3: Answer
Option (B).

Final Answer:
pi^2. \[ \boxed{\text{(B)}\ \pi^2} \]