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Evaluate the integral $$ \int\frac{a^x}{\sqrt{1-a^{2x}}}\ dx $$

In a recent question, the OP asked what the problem was with their method and why it did not match the reference book. I tried the integration in my way and arrived at a yet different answer!

My Work $$ \int\frac{a^x}{\sqrt{1-a^{2x}}}\ dx $$ Let $1-a^{2x} = u^2$ $$ \begin{align} \implies &-\color{red}{2}a^{2x}\ln a\ dx = \color{red}2\ u\ du \\ \implies &\ dx = \frac{u\ du}{-a^{2x}\ln a} \end{align} $$ Substituting in the values, I get, $$ \begin{split} \frac{-1}{\ln a}\int \frac{a^x\cdot \color{red}{u}\ du}{\color{red}{u}\cdot a^{2x}}& = \frac{-1}{\ln a} \int \frac{du}{\sqrt {1-u^2}}\\ & = \frac{- \sin ^{-1}(u)}{\ln a} \end{split} $$ Substituting in the value of $u$, I get: $$ \boxed {\color{blue}{\frac{- \sin ^{-1}(\sqrt{1-a^{2x}})}{\ln a} + C}} $$ However, the reference book has this as the answer: $$ \boxed {\color{green}{\frac{\sin ^{-1}(a^{x})}{\ln a} + C} } $$
Where have I gone wrong?

Daniele Tampieri
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Random Math Enthusiast
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1 Answers1

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Direct and fast response

The error is simply that those results are equal to less than a constant $C$.

$$-\frac{\arcsin\left(\sqrt{1-a^{2x}}\right)}{\ln\left(a\right)}+\frac{\pi}{2\ln\left(a\right)}=\frac{\arcsin\left(a^{x}\right)}{\ln\left(a\right)}$$

If in the first result you call one $C$ and in the other $C'$ you can find the relationship between $C$ and $C'$

Suggest:

Use the fact that

$$\arcsin(\sqrt{1-x^2})=\arccos(x)=-\arcsin\left(x\right)+\frac{\pi}{2}\text{ for }x\geq 0$$ and $a^x>0$ for all $x$

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