Given
$$\int\frac{1}{\sqrt {e^{4x}-1}} dx ,$$
is it possible to find the integral by $u$-substitution and using the formula: $\frac{1}{a}\tan^{-1}\left(\frac{1}{a}\right)+c$. If so, how?
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$\begingroup$ If you're looking to use trigonometric substitution, the expression of the form $\sqrt{(e^{2 x})^2 - 1}$ in the denominator suggests substituting $u = e^{2x}$, so that $du = 2 e^{2x} dx = 2 u dx$ and hence $dx = 2 \frac{du}{u}$. The obvious substitution after this point does not lead to the particular expression you mention. (Also, the "formula" you wrote is not a formula, but rather an expression.) $\endgroup$– Travis WillseFeb 27, 2016 at 18:05
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3 Answers
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Let $\displaystyle u=\frac{1}{\sqrt{e^{4x}-1}},\;\;$ so $\displaystyle du=\frac{-2e^{4x}}{(e^{4x}-1)^{3/2}}dx$.
Then $\displaystyle\int\frac{1}{\sqrt{e^{4x}-1}}dx=-\frac{1}{2}\int\frac{e^{4x}-1}{e^{4x}}\cdot\frac{-2e^{4x}}{(e^{4x}-1)^{3/2}}dx=-\frac{1}{2}\int\frac{1}{\frac{1}{e^{4x}-1}+1}\cdot\frac{-2e^{4x}}{(e^{4x}-1)^{3/2}}dx$
$\displaystyle=-\frac{1}{2}\int\frac{1}{u^2+1}du=-\frac{1}{2}\tan^{-1}u+C=-\frac{1}{2}\tan^{-1}\left(\frac{1}{\sqrt{e^{4x}-1}}\right)+C$
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There is an easy way to solve this problem without using substitution. In fact \begin{eqnarray} \int\frac{1}{\sqrt{e^{4x}-1}}dx&=&\int\frac{1}{e^{2x}\sqrt{1-e^{-4x}}}dx\\ &=&\int\frac{e^{-2x}}{\sqrt{1-e^{-4x}}}dx\\ &=&-\frac12\int\frac{1}{\sqrt{1-(e^{-2x})^2}}d(e^{-2x})\\ &=&-\frac12\arcsin(e^{-2x})+C. \end{eqnarray}
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Also, $\int \dfrac{1}{\sqrt{e^{4x}-1}} dx $=$\int \dfrac{1}{\sqrt{e^{4x}-1}} \dfrac{e^x}{e^x}$. Making the substitution $u=e^x$; $du=e^x dx$, we get
$\int \dfrac{1}{u\sqrt{u^4-1}} du$. Now, if we put $u=\sqrt {sec t}$, $du=\dfrac{sec t \cdot tg t}{2\sqrt{sec t}}$ we get $\dfrac{1}{2}\int 1 dt=\dfrac{t}{2}$ and turning back.. $\int \dfrac{1}{\sqrt{e^{4x}-1}} dx=\dfrac{sec^{-1}(e^{2x})}{2}$ This method usually works for functions involving only e^x.
