Q83 / 180 · COMEDK 2022 · Mathematics
$\int \dfrac{3^x}{\sqrt{1-9^x}} d x$ is equal to
- A$(\log 3) \sin ^{-1} 3^x+C$
- B$\dfrac{1}{3} \sin ^{-1}\left(3^x\right)+C$
- Correct answerC$\dfrac{1}{\log 3} \sin ^{-1} 3^x+C$
- D$3 \log 3 \sin ^{-1} 3^x+C$
Solution
Let $I = \int \dfrac{3^x}{\sqrt{1-9^x}} dx$.
Rewrite the denominator as $\sqrt{1-(3^x)^2}$.
Let $u = 3^x$. Then $du = 3^x \ln 3 dx$, which implies $3^x dx = \dfrac{du}{\ln 3}$.
Substituting these into the integral:
$I = \int \dfrac{1}{\sqrt{1-u^2}} \cdot \dfrac{du}{\ln 3}$
$I = \dfrac{1}{\ln 3} \int \dfrac{du}{\sqrt{1-u^2}}$
Using the standard integral formula $\int \dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1} x + C$, we get:
$I = \dfrac{1}{\ln 3} \sin^{-1} u + C$
Substituting $u = 3^x$ back:
$I = \dfrac{1}{\ln 3} \sin^{-1} 3^x + C$.
Answer: $\dfrac{1}{\log 3} \sin ^{-1} 3^x+C$
Rewrite the denominator as $\sqrt{1-(3^x)^2}$.
Let $u = 3^x$. Then $du = 3^x \ln 3 dx$, which implies $3^x dx = \dfrac{du}{\ln 3}$.
Substituting these into the integral:
$I = \int \dfrac{1}{\sqrt{1-u^2}} \cdot \dfrac{du}{\ln 3}$
$I = \dfrac{1}{\ln 3} \int \dfrac{du}{\sqrt{1-u^2}}$
Using the standard integral formula $\int \dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1} x + C$, we get:
$I = \dfrac{1}{\ln 3} \sin^{-1} u + C$
Substituting $u = 3^x$ back:
$I = \dfrac{1}{\ln 3} \sin^{-1} 3^x + C$.
Answer: $\dfrac{1}{\log 3} \sin ^{-1} 3^x+C$
Hence, the correct answer is option: 'C'.