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TomLikesPhysics
I want to integrate 2sinx/[cosx]^3. I wrote that as tanx/[cosx]^2 and saw that I now have to integrate f/f`. But somehow this does not help. Is there some formula I can use to integrate f/f`?
let cos x = u => -sin x dx = du
The given question comes out to be \[2\tan x \sec ^{2}x.dx\] which can be written as, \[\int\limits_{?}^{?}2\sec x.secx.tanx.dx\] now take secx=t, dt=secx.tanx then integrate
Hmm I just worked it out using substitution as experimentX suggested. I just thought since there is a "shortcut" to integrate f`/f as ln(f) that there is also a similar way to integrate f/f`. We have not (and will not) be introduced to secx in school so I guess it is not supposed to be the answer - also if it is right. Thx anyway cibychak and of course experimentX.
not introduced to secx?
Yup. Somehow that is not part of the mathematics-class in germany. I guess I hear about that kind of stuff when I go to university.