TS EAMCET · Maths · Definite Integration
\[ \int_0^\pi \frac{x \cos ^2 x}{1+\sin x} d x= \]
- A \(\frac{\pi(\pi-2)}{2}\)
- B 1
- C \(\frac{\pi(x+2)}{2}\)
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi(\pi-2)}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {} I=\int_0^\pi \frac{x \cos ^2 x}{1+\sin x} d x \\ & \Rightarrow I=\int_0^\pi \frac{(\pi-x) \cos ^2 x}{1+\sin x} d x \\ & \Rightarrow \quad 2 I=\int_0^\pi \frac{\pi \cos ^2 x}{1+\sin x} d x \\ & \Rightarrow 2 I=\pi \int_0^\pi \frac{1-\sin ^2 x}{1+\sin x}…
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