Q116 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
$\int_{-\pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x=$
- A$15 \pi^2 / 8$
- Correct answerB$-\pi^2 / 8$
- C$-7 \pi^2 / 8$
- D$7 \pi^2 / 8$
Solution
$I=\int_{-\pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x$
$=\int_{-\pi / 2}^{\pi / 2} \sin ^{-1}(\sin x) d x+\int_{\pi / 2}^{3 \pi / 2} \sin ^{-1}(\sin x) d x$
$+\int_{3 \pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x$
$\begin{aligned} & =\int_{-\pi / 2}^{\pi / 2} x d x+\int_{\pi / 2}^{3 \pi / 2}(\pi-x) d x+\int_{3 \pi / 2}^{2 \pi}(-2 \pi+x) d x \\ & =\left[\frac{x^2}{2}\right]_{-\pi / 2}^{\pi / 2}+\left[\pi x-\frac{x^2}{2}\right]_{\pi / 2}^{3 \pi / 2}+\left[-2 \pi x+\frac{x^2}{2}\right]_{3 \pi / 2}^{2 \pi} \\ & =0+\frac{3 \pi^2}{2}-\frac{9 \pi^2}{8}-\frac{\pi^2}{2}+\frac{\pi^2}{8}-4 \pi^2+2 \pi^2\end{aligned}$
$+3 \pi^2-\frac{9 \pi^2}{8}$
$=\pi^2+\pi^2-\frac{17 \pi^2}{8}=2 \pi^2-\frac{17 \pi^2}{8}=-\frac{\pi^2}{8}$
Hence, the correct answer is option: 'B'.