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: Use the property sin⁻¹(sin x) with periodicity.
The function is periodic with period 2π. For x ∈ [–π/2, 3π/2], it equals x on [–π/2, π/2], π–x on [π/2, 3π/2]. For [3π/2, 2π], add 2π: sin⁻¹(sin x) = x – 2π.

Step 2: Split and integrate.
∫₋π/₂^π/₂ x dx + ∫π/₂^3π/₂ (π–x) dx + ∫₃π/₂^2π (x–2π) dx.

Step 3: Evaluate each piece.
First = 0 (odd function). Second = [πx – x²/2] from π/2 to 3π/2 = 0. Third = [x²/2 – 2πx] from 3π/2 to 2π = –π²/8.

Step 4: Final Answer:
–π²/8.