AP EAMCET · Maths · Definite Integration
\(\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\)
- A \(2 \pi\)
- B \(\pi^2\)
- C \(\pi+2\)
- D \(\pi / 2\)
Answer & Solution
Correct Answer
(B) \(\pi^2\)
Step-by-step Solution
Detailed explanation
Let \(\quad I=\int_{-\pi}^\pi\left[\frac{2 x(1+\sin x)}{\left(1+\cos ^2 x\right)}\right] d x\) \(=\int_{-\pi}^\pi\left[\frac{2 x}{\left(1+\cos ^2 x\right)}\right] d x+\int_{-\pi}^\pi\) \(\left[\frac{2 x(\sin x)}{\left(1+\cos ^2 x\right)}\right] d x\) \(I=I_1+I_2\) Take…
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