If \(\int\displaystyle\frac{3^x}{\sqrt{1-9^x}}\,dx = a\sin^{-1}\left(3^x\right)+C\), then a is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{\log 3}\)
Since \(9^x = \left(3^x\right)^2\), the substitution \(t = 3^x\) gives the standard form \(\frac{1}{\sqrt{1-t^2}}\):
\(t = 3^x,\quad dt = 3^x\log 3\,dx\)
\( \displaystyle 3^x\,dx = \frac{dt}{\log 3} \)
Then
\( \displaystyle \int\frac{3^x}{\sqrt{1-9^x}}\,dx = \frac{1}{\log 3}\int\frac{dt}{\sqrt{1-t^2}} \)
\( \displaystyle = \frac{1}{\log 3}\sin^{-1}\left(3^x\right)+C \)
So \(a = \frac{1}{\log 3}\).
Why the other options are incorrect
Option (1) \(\log 3\) — Incorrect. The derivative of \(3^x\) is \(3^x\log 3\), so the integral is divided by \(\log 3\), not multiplied.
Option (2) \(\frac{1}{2}\) — Incorrect. No factor 2 arises; \(9^x\) is simply the square of \(3^x\).
Option (3) 1 — Incorrect. It ignores the factor \(\log 3\) from \(dt = 3^x\log 3\,dx\).