Question

Evaluate : $\int_{- \pi/2}^{2\pi}\sin^{- 1}$ (sin x) dx -

• A. – $\frac{\pi^{2}}{4}$
• B. $\frac{–\pi^{2}}{8}$
• C. $\frac{\pi^{2}}{4}$
• D. None of these

Answer 1

Answer: (B)

$\frac{–\pi^{2}}{8}$

Answer 2

The graph of ƒ(x) = sin^–1 (sin x) is shown as in figure.

Therefore$\int_{- \pi/2}^{2\pi}\sin^{- 1}$ (sin x) dx = $\int_{- \pi/2}^{0}\sin^{- 1}$ (sin x) dx+ $\int_{0}^{\pi}\sin^{- 1}$ (sin

x) dx + $\int_{\pi}^{2\pi}\sin^{- 1}$ (sin x) dx

= Area of shaded region

= – $\left( \frac{1}{2} \times \frac{\pi}{2} \times \frac{\pi}{2} \right)$ + $\left( \frac{1}{2} \times \pi \times \frac{\pi}{2} \right)$ – $\left( \frac{1}{2} \times \pi \times \frac{\pi}{2} \right)$

= – $\frac{\pi^{2}}{8}$.