Does anyone know how to integrate: I've tried doing it in terms of e and I haven't really got anywhere. Please help!
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Originally Posted by Showcase_22 Does anyone know how to integrate: I've tried doing it in terms of e and I haven't really got anywhere. Please help! You can multiply it with $\displaystyle \frac{tanh(x)}{tanh(x)}$ Remember that: $\displaystyle (\text{sech(x)})' = -\text{sech(x)}\text{tanh(x)}$ $\displaystyle tanh(x) = \sqrt{1-sech^2(x)}$ By substituting for u = sech(x), you will get a standard form in the end.
Originally Posted by Showcase_22 Does anyone know how to integrate: I've tried doing it in terms of e and I haven't really got anywhere. Please help! $\displaystyle = \int \frac{2}{e^x + e^{-x}} \, dx$ and the substitution $\displaystyle u = e^x$ works just fine: $\displaystyle \int \frac{2}{u^2 + 1} \, du \, ....$
oh right. I think I tried turning it into partial fractions which is a very long and arduous process. cheers!
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