Question:
\(\displaystyle \int \frac{3\,dx}{\sqrt{\,1-9x^{2}\,}}=\) ?
\(\displaystyle \int \frac{3\,dx}{\sqrt{\,1-9x^{2}\,}}=\) ?
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Standard form: $\displaystyle \int\frac{du}{\sqrt{1-u^{2}}}=\sin^{-1}u+C$.
Updated On: Oct 17, 2025
- \(\tan^{-1}(3x)+k\)
- \(\sec^{-1}(3x)+k\)
- \(\sin^{-1}(3x)+k\)
- \(\cos^{-1}(3x)+k\)
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The Correct Option is C
Solution and Explanation
Let \(u=3x\Rightarrow du=3\,dx\). Then
\[
\int \frac{3\,dx}{\sqrt{1-9x^{2}}}=\int \frac{du}{\sqrt{1-u^{2}}}=\sin^{-1}u+k=\sin^{-1}(3x)+k.
\]
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