Question

\(\displaystyle \int \frac{3\,dx}{\sqrt{\,1-9x^{2}\,}}=\) ?

• A. \(\tan^{-1}(3x)+k\)
• B. \(\sec^{-1}(3x)+k\)
• C. \(\sin^{-1}(3x)+k\)
• D. \(\cos^{-1}(3x)+k\)

Hint:

Standard form: $\displaystyle \int\frac{du}{\sqrt{1-u^{2}}}=\sin^{-1}u+C$.

Answer 1

The Correct Option is C

Let \(u=3x\Rightarrow du=3\,dx\). Then \[ \int \frac{3\,dx}{\sqrt{1-9x^{2}}}=\int \frac{du}{\sqrt{1-u^{2}}}=\sin^{-1}u+k=\sin^{-1}(3x)+k. \]