Question

\[ \int_{-1}^1 \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx \text{ is equal to} \]

• A. 1
• B. 0
• C. 4
• D. 2

Hint:

Nuclear reactions must conserve both mass and charge.

Answer 1

The Correct Option is D

Let us denote \[ I = \int_{-1}^1 \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx \] This is an odd function times an odd function over symmetric limits, thus the integrand is even. Use substitution: Let \(x = \sin \theta\), then \[ dx = \cos \theta d\theta,\quad \sqrt{1 - x^2} = \cos \theta,\quad \sin^{-1} x = \theta \] \[ I = \int_{x=-1}^1 \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx = \int_{\theta = -\pi/2}^{\pi/2} \sin \theta \cdot \theta d\theta = 2 \int_0^{\pi/2} \theta \sin \theta d\theta \] Use integration by parts: \[ = 2\left[-\theta \cos \theta + \int \cos \theta d\theta \right]_0^{\pi/2} = 2 \left[-\theta \cos \theta + \sin \theta\right]_0^{\pi/2} = 2(0 + 1 - 0) = 2 \]