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integral e^(-2x)tan(e^-2x)dx please give me some hints...
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\[\huge \int\limits e^{-2 x} \left(\tan [e^{-2 x}]\right) \, dx\] Try u-sub
u=-2x du=-2dx \[-\frac{du}{2}=dx\] \[\large =\int\limits e^u(\tan(e^u))(-\frac{du}{2})\] \[\large =-\frac{1}{2} \int\limits \left(e^u \tan (e^u)\right) \, du\]
very detailed, thank u so much ^^
yw :)
Hmm wouldn't you rather set the entire exponential as your u? I mean unless you don't mind doing a second substitution.
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can i let u=e^-2x then it will be \[\int\limits -\frac{ 1 }{ 2} \tan u du\] is it right?
Yah that looks good! :)
tysm ^^
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