Integration question (equation in comment)

Question

anonymous · Answer

$$\int\limits_{a}^{b}\frac{ 6e ^{x} }{\sqrt{1-e ^{2x}}}dx$$

anonymous · Answer

Did you consider a $u$ substitution $u=e^x$ such that $du = e^xdx=udx$
so your integral would look like this: $$\large \int\frac{6u}{\sqrt{1-u^2}} \frac{1}{u}du \implies 6\int \frac{1}{\sqrt{1-u^2}}du $$

anonymous · Answer

the integral on the right should ring a bell. If not, you can continue to play from there with trigonometric substitutions.

anonymous · Answer

I know the integral on the right is equal to $$\sin^{-1} (u)$$ but that's in terms of u. How would I get an answer in terms of x?

anonymous · Answer

well, you just back substitute, remember the following rule of thumb "Every substitution, requires a back-substition"

anonymous · Answer

so your answer so far is $\sin^{-1}(u)$ and you know that $u=e^x$ so your final answer would be?

anonymous · Answer

But if the integral were equal to $$6\sin^{-1} (e ^{x})$$ then wouldn't its integrand be $$6\int\limits_{a}^{b} [\frac{ 1 }{ \sqrt{1-e ^{x}} }*e ^{x}]dx$$ because of the chain rule?

anonymous · Answer

Oops, *it be, not 'its integrand be'

anonymous · Answer

you almost did it correct, look at the denominator though, where is your flaw?
The derivative of $\sin^{-1}(x)$ is not quite what you have posted above.

anonymous · Answer

Oh, yeah, it'd be $$e ^{2x} $$ in the denominator. Thank you!

anonymous · Answer

you're very welcome.