Q114 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

$$ \int_0^\pi \frac{x \cos ^2 x}{1+\sin x} d x= $$

  • Correct answer
    A
    $\frac{\pi(\pi-2)}{2}$
  • B
    1
  • C
    $\frac{\pi(x+2)}{2}$
  • D
    $\frac{\pi}{4}$

Solution

$\begin{aligned} & \text {} I=\int_0^\pi \frac{x \cos ^2 x}{1+\sin x} d x \\ & \Rightarrow I=\int_0^\pi \frac{(\pi-x) \cos ^2 x}{1+\sin x} d x \\ & \Rightarrow \quad 2 I=\int_0^\pi \frac{\pi \cos ^2 x}{1+\sin x} d x \\ & \Rightarrow 2 I=\pi \int_0^\pi \frac{1-\sin ^2 x}{1+\sin x} d x \\ & \Rightarrow I=\frac{\pi}{2} \int_0^\pi(1-\sin x) d x \\ & \Rightarrow I=\frac{\pi}{2}[x+\cos x]_0^\pi \\ & =\frac{\pi}{2}(\pi-1-1)=\frac{\pi(\pi-2)}{2}\end{aligned}$

Hence, the correct answer is option: 'A'.

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