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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`?
 one year ago
 one year ago
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`?
 one year ago
 one year ago

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experimentXBest ResponseYou've already chosen the best response.1
let cos x = u => sin x dx = du
 one year ago

cibychakBest ResponseYou've already chosen the best response.1
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
 one year ago

TomLikesPhysicsBest ResponseYou've already chosen the best response.0
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.
 one year ago

cibychakBest ResponseYou've already chosen the best response.1
not introduced to secx?
 one year ago

TomLikesPhysicsBest ResponseYou've already chosen the best response.0
Yup. Somehow that is not part of the mathematicsclass in germany. I guess I hear about that kind of stuff when I go to university.
 one year ago
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