If 0 le x le 1, I_1 = int sin^{-1} sqrt{1 - x | vedclass

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(D) Given $I_1 = \int \sin^{-1} \sqrt{1 - x^2} dx$ and $I_2 = \int \sin^{-1} x dx$.Since $\sin^{-1} \sqrt{1 - x^2} = \cos^{-1} x$ for $0 \le x \le 1$,

Vedclass · Accepted Answer

$I_1 + I_2 = x \frac{\pi}{2}$

Vedclass · Answer

(D) Given $I_1 = \int \sin^{-1} \sqrt{1 - x^2} dx$ and $I_2 = \int \sin^{-1} x dx$.Since $\sin^{-1} \sqrt{1 - x^2} = \cos^{-1} x$ for $0 \le x \le 1$, we have $I_1 = \int \cos^{-1} x dx$.Now, $I_1 + I_2 = \int (\sin^{-1} x + \cos^{-1} x) dx$.Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, we get $I_1 + I_2 = \int \frac{\pi}{2} dx$.Therefore, $I_1 + I_2 = \frac{\pi}{2} x + C$.

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

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