int frac{a^x}{sqrt{1 - a^{2x}}} dx = | vedclass

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Question

(A) Let $a^x = t$. Then,differentiating both sides with respect to $x$,we get $a^x \log_e a \, dx = dt$,which implies $a^x \, dx = \frac{dt}{\log_e a}

Vedclass · Accepted Answer

$\frac{1}{\log a} \sin^{-1}(a^x) + c$

Vedclass · Answer

(A) Let $a^x = t$. Then,differentiating both sides with respect to $x$,we get $a^x \log_e a \, dx = dt$,which implies $a^x \, dx = \frac{dt}{\log_e a}$.Substituting these into the integral:$\int \frac{a^x}{\sqrt{1 - a^{2x}}} dx = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{dt}{\log_e a}$$= \frac{1}{\log_e a} \int \frac{dt}{\sqrt{1 - t^2}}$$= \frac{1}{\log_e a} \sin^{-1}(t) + c$$= \frac{\sin^{-1}(a^x)}{\log_e a} + c$.

$\int \frac{a^x}{\sqrt{1 - a^{2x}}} dx = $

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