COMEDK · Maths · 28. Indefinite Integration
\(\int \dfrac{4^x}{\sqrt{1-16^x}} d x\) is equal to
- A \((\log 4) \sin ^{-1} 4^x+C\)
- B \(4 \log 4 \sin ^{-1} 4+C\)
- C \(\dfrac{1}{4} \sin ^{-1}\left(4^x\right)+C\)
- D \(\dfrac{1}{\log 4} \sin ^{-1} 4^x+C\)
Answer & Solution
Correct Answer
(D) \(\dfrac{1}{\log 4} \sin ^{-1} 4^x+C\)
Step-by-step Solution
Detailed explanation
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…
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