Q119 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
$\int \dfrac{4^x}{\sqrt{1-16^x}} d x$ is equal to
- A$(\log 4) \sin ^{-1} 4^x+C$
- B$4 \log 4 \sin ^{-1} 4+C$
- C$\dfrac{1}{4} \sin ^{-1}\left(4^x\right)+C$
- Correct answerD$\dfrac{1}{\log 4} \sin ^{-1} 4^x+C$
Solution
Let $I = \int \dfrac{4^x}{\sqrt{1-16^x}} dx$.
Rewrite the integral as $I = \int \dfrac{4^x}{\sqrt{1-(4^x)^2}} dx$.
Substitute $u = 4^x$. Then $du = 4^x \ln 4 dx$, which implies $4^x dx = \dfrac{du}{\ln 4}$.
Substituting these into the integral gives $I = \int \dfrac{1}{\sqrt{1-u^2}} \cdot \dfrac{du}{\ln 4}$.
$I = \dfrac{1}{\ln 4} \int \dfrac{du}{\sqrt{1-u^2}}$.
Using the standard integral $\int \dfrac{du}{\sqrt{1-u^2}} = \sin^{-1} u + C$, we get $I = \dfrac{1}{\ln 4} \sin^{-1} u + C$.
Substituting back $u = 4^x$, we obtain $I = \dfrac{1}{\ln 4} \sin^{-1} 4^x + C$.
Answer: $\dfrac{1}{\log 4} \sin ^{-1} 4^x+C$
Rewrite the integral as $I = \int \dfrac{4^x}{\sqrt{1-(4^x)^2}} dx$.
Substitute $u = 4^x$. Then $du = 4^x \ln 4 dx$, which implies $4^x dx = \dfrac{du}{\ln 4}$.
Substituting these into the integral gives $I = \int \dfrac{1}{\sqrt{1-u^2}} \cdot \dfrac{du}{\ln 4}$.
$I = \dfrac{1}{\ln 4} \int \dfrac{du}{\sqrt{1-u^2}}$.
Using the standard integral $\int \dfrac{du}{\sqrt{1-u^2}} = \sin^{-1} u + C$, we get $I = \dfrac{1}{\ln 4} \sin^{-1} u + C$.
Substituting back $u = 4^x$, we obtain $I = \dfrac{1}{\ln 4} \sin^{-1} 4^x + C$.
Answer: $\dfrac{1}{\log 4} \sin ^{-1} 4^x+C$
Hence, the correct answer is option: 'D'.