Question:
\(\int_0^1 \sin^{-1}x \, dx\) is
\(\int_0^1 \sin^{-1}x \, dx\) is
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Inverse trig integrals → use integration by parts.
Updated On: Apr 18, 2026
- \(\pi/2 -1\)
- 1
- \(\pi\)
- 0
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The Correct Option is A
Solution and Explanation
Concept:
Use integration by parts.
Step 1: Let
\[ u=\sin^{-1}x,\quad dv=dx \]
Step 2: Apply formula.
\[ \int u\,dv = uv - \int v\,du \] \[ = x\sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}}dx \]
Step 3: Evaluate.
\[ = x\sin^{-1}x + \sqrt{1-x^2} \]
Step 4: Apply limits.
\[ = \left[\frac{\pi}{2} -1\right] \]
Step 1: Let
\[ u=\sin^{-1}x,\quad dv=dx \]
Step 2: Apply formula.
\[ \int u\,dv = uv - \int v\,du \] \[ = x\sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}}dx \]
Step 3: Evaluate.
\[ = x\sin^{-1}x + \sqrt{1-x^2} \]
Step 4: Apply limits.
\[ = \left[\frac{\pi}{2} -1\right] \]
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