Question:
\( \int \frac{\sin^{-1}x}{\sqrt{1-x^2}} \, dx = \)
\( \int \frac{\sin^{-1}x}{\sqrt{1-x^2}} \, dx = \)
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If derivative of inverse trig appears, use substitution directly.
Updated On: Apr 21, 2026
- \( \frac{1}{2}(\sin^{-1}x)^2 + C \)
- \( -(\sin^{-1}x)\sqrt{1-x^2} + C \)
- \( (\sin^{-1}x)\sqrt{1-x^2} + x + C \)
- \( (\sin^{-1}x)\sqrt{1-x^2} - x + C \)
- \( (\sin^{-1}x)^2 + C \)
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The Correct Option is A
Solution and Explanation
Concept:
Let \( t = \sin^{-1}x \Rightarrow dt = \frac{1}{\sqrt{1-x^2}}dx \)
Step 1: Substitute.
\[ \int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx = \int t\,dt \]
Step 2: Integrate.
\[ = \frac{t^2}{2} + C = \frac{1}{2}(\sin^{-1}x)^2 + C \]
Step 1: Substitute.
\[ \int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx = \int t\,dt \]
Step 2: Integrate.
\[ = \frac{t^2}{2} + C = \frac{1}{2}(\sin^{-1}x)^2 + C \]
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