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`?

Question

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`?

experimentx · Answer

let cos x = u
=> -sin x dx = du

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

not introduced to secx?

experimentx · Answer

yw

anonymous · Answer

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.

anonymous · Answer

oh :)