Q85 / 180 · COMEDK 2021 · Mathematics
$\int \dfrac{2^{x}}{\sqrt{1-4^{x}}} d x$ is equal to
- A$(\log 2) \sin ^{-1} 2^{x}+C$
- B$\dfrac{1}{2} \sin ^{-1} 2^{x}+C$
- Correct answerC$\dfrac{1}{\log 2} \sin ^{-1} 2^{x}+C$
- D$2 \log 2 \sin ^{-1} 2^{x}+C$
Solution
Let $I = \int \dfrac{2^{x}}{\sqrt{1-4^{x}}} dx$.
Rewrite the integral as $I = \int \dfrac{2^{x}}{\sqrt{1-(2^{x})^{2}}} dx$.
Substitute $u = 2^{x}$. Then $du = 2^{x} \ln 2 \, dx$, which implies $2^{x} dx = \dfrac{du}{\ln 2}$.
Substituting these into the integral gives $I = \int \dfrac{1}{\sqrt{1-u^{2}}} \cdot \dfrac{du}{\ln 2}$.
$I = \dfrac{1}{\ln 2} \int \dfrac{du}{\sqrt{1-u^{2}}}$.
Using the standard integral formula $\int \dfrac{du}{\sqrt{1-u^{2}}} = \sin^{-1} u + C$, we get $I = \dfrac{1}{\ln 2} \sin^{-1} u + C$.
Substituting back $u = 2^{x}$, we obtain $I = \dfrac{1}{\ln 2} \sin^{-1} 2^{x} + C$.
Answer: $\dfrac{1}{\log 2} \sin ^{-1} 2^{x}+C$
Rewrite the integral as $I = \int \dfrac{2^{x}}{\sqrt{1-(2^{x})^{2}}} dx$.
Substitute $u = 2^{x}$. Then $du = 2^{x} \ln 2 \, dx$, which implies $2^{x} dx = \dfrac{du}{\ln 2}$.
Substituting these into the integral gives $I = \int \dfrac{1}{\sqrt{1-u^{2}}} \cdot \dfrac{du}{\ln 2}$.
$I = \dfrac{1}{\ln 2} \int \dfrac{du}{\sqrt{1-u^{2}}}$.
Using the standard integral formula $\int \dfrac{du}{\sqrt{1-u^{2}}} = \sin^{-1} u + C$, we get $I = \dfrac{1}{\ln 2} \sin^{-1} u + C$.
Substituting back $u = 2^{x}$, we obtain $I = \dfrac{1}{\ln 2} \sin^{-1} 2^{x} + C$.
Answer: $\dfrac{1}{\log 2} \sin ^{-1} 2^{x}+C$
Hence, the correct answer is option: 'C'.