Question

If \(0\leq x\leq 1\), \(I_1 = \int sin^{-1}\sqrt{1-x^2}\,dx\) and \(I_2 = \int sin^{-1}x\,dx\), then which of the following is true?

• A. \(I_1 = I_2\)
• B. \(I_1 = \frac{π}{2}I_2\)
• C. \(I_1+I_2 = \frac{π}{2}x\)
• D. \(I_1+I_2 = \frac{π}{2}\)

Hint:

For x in [0,1], arcsin of the root of one minus x squared is arccos x.

Answer 1

The Correct Option is C

Step 1: Rewrite $I_1$:
Because $\sin^{-1}\sqrt{1 - x^2} = \cos^{-1}x$ on $[0,1]$, $I_1$ is the integral of $\cos^{-1}x$.

Step 2: Add:
The sum of integrands is $\frac\pi2$, a constant, so the integral is $\frac\pi2 x$ (ignoring the constant of integration).

Final Answer:
$I_1 + I_2 = \frac{\pi}{2}x$, option (C). \[ \boxed{I_1 + I_2 = \frac{\pi}{2}x} \]