Integral Calculus: ∫dm/√e^-4m -1

Question

anonymous · Answer

can someone please tell me how to start? 
This is the clearer version:
$$\int\limits_{}^{} \frac{ dm }{ \sqrt{e ^{-4m}-1}}$$

anonymous · Answer

After some simplifications, we get
$$\int\limits_{}^{} \frac{ e^{2m} dm }{ \sqrt{1-e^{2m}}} $$

anonymous · Answer

Sorry for asking this, but how did you simplify it?

anonymous · Answer

Put
$$e^{-4m} = \frac{ 1 }{ e^{4m} }$$

anonymous · Answer

Oh. I see. Thanks!

anonymous · Answer

HEy sorry it's a mistake there..

anonymous · Answer

$$\int\limits_{}^{}\frac{ e^{2m} dm }{ \sqrt{1-e^{4m}} }$$

anonymous · Answer

Wait. After simplification, does this mean that I can now proceed in using the inverse trigonometric formula for arcsin? $$u ^{2}=e ^{2m}$$ but the derivative of u will not be equal to the one on the top. there is still an e. Um what will happen? Thanks!

anonymous · Answer

put e^2m = u

anonymous · Answer

then du/2 = e^2m dm

anonymous · Answer

$$\frac{1}{2} \int\limits_{}^{} \frac{ du }{ \sqrt{1-u^2} }$$

anonymous · Answer

Thank you very much! :) I'm sorry, I didn't see the corrected version.

anonymous · Answer

No problem.