Question

\(\displaystyle \int_{-\pi/2}^{2\pi}\sin^{-1}(\sin x)\,dx=\)

• A. \(\dfrac{15\pi^2}{8}\)
• B. \(-\dfrac{\pi^2}{8}\)
• C. \(-\dfrac{7\pi^2}{8}\)
• D. \(\dfrac{7\pi^2}{8}\)

Hint:

For \(\sin^{-1}(\sin x)\), always convert it into its piecewise form according to the principal range \[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. \] This avoids mistakes in definite integration.

Answer 1

The Correct Option is B

Step 1: Write the piecewise form of \(\sin^{-1}(\sin x)\).
The function \(\sin^{-1}(\sin x)\) always gives values in \[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \] For the interval \[ -\frac{\pi}{2}\leq x\leq 2\pi, \] we write \[ \sin^{-1}(\sin x)=x,\qquad -\frac{\pi}{2}\leq x\leq \frac{\pi}{2} \] \[ \sin^{-1}(\sin x)=\pi-x,\qquad \frac{\pi}{2}\leq x\leq \frac{3\pi}{2} \] \[ \sin^{-1}(\sin x)=x-2\pi,\qquad \frac{3\pi}{2}\leq x\leq 2\pi \]

Step 2: Split the integral.
Let \[ I=\int_{-\pi/2}^{2\pi}\sin^{-1}(\sin x)\,dx \] Then, \[ I= \int_{-\pi/2}^{\pi/2}x\,dx + \int_{\pi/2}^{3\pi/2}(\pi-x)\,dx + \int_{3\pi/2}^{2\pi}(x-2\pi)\,dx \]

Step 3: Evaluate the first integral.
\[ \int_{-\pi/2}^{\pi/2}x\,dx = \left[\frac{x^2}{2}\right]_{-\pi/2}^{\pi/2} \] \[ = \frac{\pi^2}{8}-\frac{\pi^2}{8} \] \[ =0 \]

Step 4: Evaluate the second integral.
\[ \int_{\pi/2}^{3\pi/2}(\pi-x)\,dx = \left[\pi x-\frac{x^2}{2}\right]_{\pi/2}^{3\pi/2} \] At \(x=\dfrac{3\pi}{2}\), \[ \pi x-\frac{x^2}{2} = \frac{3\pi^2}{2}-\frac{9\pi^2}{8} = \frac{3\pi^2}{8} \] At \(x=\dfrac{\pi}{2}\), \[ \pi x-\frac{x^2}{2} = \frac{\pi^2}{2}-\frac{\pi^2}{8} = \frac{3\pi^2}{8} \] Thus, \[ \int_{\pi/2}^{3\pi/2}(\pi-x)\,dx=0 \]

Step 5: Evaluate the third integral.
\[ \int_{3\pi/2}^{2\pi}(x-2\pi)\,dx = \left[\frac{x^2}{2}-2\pi x\right]_{3\pi/2}^{2\pi} \] At \(x=2\pi\), \[ \frac{x^2}{2}-2\pi x = 2\pi^2-4\pi^2 = -2\pi^2 \] At \(x=\dfrac{3\pi}{2}\), \[ \frac{x^2}{2}-2\pi x = \frac{9\pi^2}{8}-3\pi^2 = -\frac{15\pi^2}{8} \] Therefore, \[ \int_{3\pi/2}^{2\pi}(x-2\pi)\,dx = -2\pi^2+\frac{15\pi^2}{8} \] \[ = -\frac{\pi^2}{8} \]

Step 6: Final conclusion.
Hence, \[ I=0+0-\frac{\pi^2}{8} \] \[ \boxed{-\frac{\pi^2}{8}} \]