If \(\int \frac{3^{x}}{\sqrt{1-9^{x}}}dx=a\sin^{-1}(3^{x})+c\), then \(a\) is equal to

\(\log 3\) \(\frac{1}{\log 3}\) \(\frac{1}{2}\) \(1\)

\(\frac{1}{\log 3}\)

Let \(3^{x}=t\) then \(\begin{aligned}\int \frac{3^{x}}{\sqrt{1-9^{x}}}dx&=\frac{1}{\log_{e}3}\int \frac{1}{\sqrt{1-t^{2}}}dt\\ &=\frac{1}{\log_{e} 3}\sin^{-1}t+c\end{aligned}\)