## anonymous 4 years 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?

1. experimentX

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

2. anonymous

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

3. anonymous

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.

4. anonymous

not introduced to secx?

5. experimentX

yw

6. anonymous

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.

7. anonymous

oh :)