Integrate $\int \frac{8}{16-e^{4x}} \mathrm dx$ using trigonometric substitution

Question

How do I integrate this equation using trigonometric substitution?

$$\int \frac{8}{16-e^{4x}} \mathrm dx.$$

So far I figured out that $a = 4, u = e^{2x}$ then $e^{2x}=4\sin\theta \mathrm d\theta$ and $\sqrt{16-e^{4x}}$ and $2e^{2x}dx=4\sin\theta$. I don't know how to proceed from this because I don't know if $e^{2x}$ can be considered a constant and can be factored out of the integral or I have to use logarithmic functions.

Edit: I tried solving it, I had a final answer of
$$\frac{e^{2x}}{4\sqrt{16-e^{4x}}}+C.$$
Am I correct?

Answer 1

$$ \int \frac{8}{16-e^{4x}} \, dx=\left| u=e^{2x} \atop du=2e^{2x}\,dx=2u\, dx \right| =\int\frac{8}{16-u^2} \cdot \frac{du}{2u} $$ (we assume that $u\in(-4,4)$) $$ =\left| u=4\sin t \atop du=4\cos t\,dt \right|= \int\frac{8}{16-16\sin^2t} \cdot \frac{4\cos t\,dt}{8\sin t} =\frac14 \int \frac{\cos t\,dt}{\cos^2t \sin t} $$ $$ =\frac12 \int \frac{dt}{2\sin t\cos t}= \frac12 \int \frac{dt}{\sin 2t}= \left| z= 2t \atop dz= 2\,dt \right|= $$ $$ =\frac14 \int \frac{dz}{\sin z}= \frac14\ln\left| \tan\left( \frac{z}2 \right) \right|+C =\frac14\ln\left| \tan t \right|+C $$ $$ =\frac14\ln\left| \tan\left( \arcsin\frac{u}4 \right) \right|+C =\frac14\ln\left| \tan\left( \arcsin\frac{e^{2x}}4 \right) \right|+C $$ Since $$ \tan^2(\arcsin x)=\frac1{\cot^2(\arcsin x)}= \frac1{\frac1{\sin^2(\arcsin x)}-1}= \frac1{\frac1{x^2}-1}=\frac{x^2}{1-x^2} $$ and (considering the sign) $$ \tan(\arcsin x)= \frac{x}{\sqrt{1-x^2}} $$ the answer is $$ \frac14\ln\left| \frac{e^{2x}/4}{1-\sqrt{\frac{e^{4x}}{16}}} \right|+C =\frac{x}2-\frac14\ln\left| \sqrt{16-e^{4x}} \right|+C $$ $$ =\frac{x}2-\frac18\ln\left| 16-e^{4x} \right|+C. $$