int frac{3^x}{sqrt{1-9^x}} d x= | vedclass

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Question

(A) Let $I = \int \frac{3^x}{\sqrt{1-9^x}} d x$.
We can rewrite the denominator as $9^x = (3^x)^2$,so $I = \int \frac{3^x}{\sqrt{1-(3^x)^2}} d x$.
S

Vedclass · Accepted Answer

$\sin ^{-1}\left(3^x\right) \cdot(\log 3)^{-1}+c$

Vedclass · Answer

(A) Let $I = \int \frac{3^x}{\sqrt{1-9^x}} d x$.
We can rewrite the denominator as $9^x = (3^x)^2$,so $I = \int \frac{3^x}{\sqrt{1-(3^x)^2}} d x$.
Substitute $t = 3^x$. Then,differentiating with respect to $x$,we get $dt = 3^x \log 3 \, dx$,which implies $3^x \, dx = \frac{1}{\log 3} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{\sqrt{1-t^2}} \cdot \frac{1}{\log 3} dt$.
Since $\int \frac{1}{\sqrt{1-t^2}} dt = \sin^{-1}(t) + c$,we have $I = \frac{1}{\log 3} \sin^{-1}(t) + c$.
Finally,substituting $t = 3^x$ back,we get $I = \frac{1}{\log 3} \sin^{-1}(3^x) + c$ or $\sin^{-1}(3^x) \cdot (\log 3)^{-1} + c$.

$\int \frac{3^x}{\sqrt{1-9^x}} d x=$

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