integrate 6 / [ cos x(2-sinx) ] dx?

Question

how do i do this can i take u = (2-sinx) and do it by substitution and cancelling the cos x at the bottom since the derivative of (1-sinx) = (-cosx),, and then integrate in terms of u

which will give me 6 ln (2-sinx) {this is not the answer on my book though }

i cant seem to get the answer can someone help,, or give me the steps how to do it

and why is my method wrong????

Answer 1

Im trying to figure it out but I see ur mistake.
You cant cancel the cos x because they are BOTH in the denominator.
∫6 / [ (cos x)(2-sinx) ] dx
u = (2 - sinx)
du = -cosx dx

∫6 / [ (cos x)(u) ] du/(-cos x)
It looks like you can cancel, however the first cos(x) is below the 6 and the second cos(x) is below the du.

EDIT: I tried but i cant figure it out even after looking at the table integrals and trig identities.

Answer 2

I loved this question. Took me an hour to solve. I tried simplifying the trigo terms, integrated by parts, and tried various methods. Finally I managed to solve it.

The answer is the link below:
http://www.rvhs.moe.edu.sg/personal/s9141057e/integral.doc

It's in Microsoft Word format because I really had to type out the working neatly (typing in here would be a disaster).

If you can't view it with Word or you encounter problems, you can download the .htm files.
http://www.rvhs.moe.edu.sg/personal/s9141057e/integral.zip

They should appear _fine_ in a web browser.

Very challenging question.

EDIT: I forgot to multiply everything by 6. But after all, it's a constant (: