Check my integration by substitution answer

[hank]

Ok, here's the problem:

\(\displaystyle \int_{\ln 2}^{\ln \frac{2}{{\sqrt 3 }}} {\frac{{e^{ - x} }}{{\sqrt {1 - e^{ - 2x} } }}}\)

Here's my answer:

\(\displaystyle - 2\ln \left( {\frac{1}{{2 - \sqrt 3 }}} \right)\)

Am I right?

 

[stapel]

I don't think so; The Integrator and I are coming up with stuff involving inverse-trig functions. How did you get your answer?

Please show all of your steps. Thank you.

Eliz.

 

[galactus]

No, your answer is not correct. Actually, the answer is much more simplistic than that.

\(\displaystyle \L\\\int_{ln(2)}^{ln(\frac{2}{\sqrt{3}})}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}dx\)


Let \(\displaystyle \L\\u=e^{-x}; \;\ du=-e^{-x}dx; \;\ u^{2}=e^{-2x}\)

Stapel is correct, you're indefinite integral is an inverse trig.

\(\displaystyle \L\\-\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{1}{\sqrt{1-u^{2}}}du\)

 

[hank]

Ok, I see where my mistake was.

I didn't know that I should raise u to the 2d power for dealing with the e^-2x.

Thanks for the help!