Q76 / 180 · COMEDK 2024 (Memory Based) · Mathematics
$\int \frac{9^x}{\sqrt{1-81^x}} d x$ is equal to
- A$\sin ^{-1}\left(9^x\right)+c$
- B$\ln ^9 \sin ^{-1}\left(9^x\right)+c$
- C$\sin ^{-1}\left(81^x\right)+c$
- Correct answerD$\frac{1}{\ln 9} \sin ^{-1}\left(9^x\right)+c$
Solution
Let $9^x=t$
Differentiate both side w.r t. $x$
$\begin{aligned}<br/>9^x \ln 9 d x & =d t \\<br/>\Rightarrow \quad 9^x d x & =\frac{d t}{\ln 9}<br/>\end{aligned}$
$\begin{aligned} & \text { Now, } \int \frac{9^x}{\sqrt{1-9^{2 x}}} d x=\frac{1}{\ln 9} \int \frac{d t}{\sqrt{1-t^2}} \\ & \quad=\frac{1}{\ln 9} \sin ^{-1}(t)+c\left[\because \int \frac{1}{\sqrt{1-t^2}} d t=\sin ^{-1} t\right] \\ & \quad=\frac{1}{\ln 9} \sin ^{-1}\left(9^x\right)+c\end{aligned}$
Differentiate both side w.r t. $x$
$\begin{aligned}<br/>9^x \ln 9 d x & =d t \\<br/>\Rightarrow \quad 9^x d x & =\frac{d t}{\ln 9}<br/>\end{aligned}$
$\begin{aligned} & \text { Now, } \int \frac{9^x}{\sqrt{1-9^{2 x}}} d x=\frac{1}{\ln 9} \int \frac{d t}{\sqrt{1-t^2}} \\ & \quad=\frac{1}{\ln 9} \sin ^{-1}(t)+c\left[\because \int \frac{1}{\sqrt{1-t^2}} d t=\sin ^{-1} t\right] \\ & \quad=\frac{1}{\ln 9} \sin ^{-1}\left(9^x\right)+c\end{aligned}$
Hence, the correct answer is option: 'D'.