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: Understanding the Concept:
For \(0 \leq x \leq 1\), put \(x = \cos\theta\) with \(\theta \in [0, \frac\pi2]\). Then \(\sqrt{1 - x^2} = \sin\theta\) and \(\sin^{-1}\sqrt{1 - x^2} = \theta = \cos^{-1}x\).

Step 2: Key Formula or Approach:
\(\sin^{-1}x + \cos^{-1}x = \frac\pi2\).

Step 3: Detailed Explanation:
So \(I_1 = \int \cos^{-1}x\,dx\) and \(I_2 = \int \sin^{-1}x\,dx\).
\[ I_1 + I_2 = \int (\cos^{-1}x + \sin^{-1}x)\,dx = \int \frac\pi2\,dx = \frac\pi2 x \]
The options A and B would need \(\cos^{-1}x\) and \(\sin^{-1}x\) to be equal or proportional, which they are not. Option D would give a constant, but the integral gives a term in \(x\).

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