Question

integral of cos x / (2 - cos x) dx

Answer 1

please give me the complete solution .

Answer 2

\[\int\limits_{}^{} \cos x / (2 - \cos x) dx\]

Answer 3

\[ {\cos x \over 2 - \cos x} = {}\left( {2\over 2 -\cos x} - 1 \right)\]
now use Weierstrass substitution

Answer 4

http://en.wikipedia.org/wiki/Weierstrass_substitution#Second_example:_a_definite_integral

Answer 5

not much need actually, after you split it as @hash.nuke did, you will find main problem is to integrate 2/(2-cosx) or 1/(1- cos x)

=> 1/(2 sin^2 x/2) = 1/2 * (cosec ^2 (x/2)) ,,hope rest is easier for you..

Answer 6

\[\frac{2}{2-\cos x} \neq \frac{1}{1-\cos x}\] @shubhamsrg

Answer 7

use this :
let tan(x/2) = t --> 1/2 * sec^2 (x/2) dx = dt --> dx=2*1/(t^2+1)*dt
cos^2(x/2) = 1/sqrt(1+t^2)
cosx = (1-t^2)/(t^2+1)
so, it can be
int {[(1-t^2)/(t^2+1)]/[(2 - (1-t^2)/(t^2+1]} * 2/(t^2+1)*dt
= int [(2-2t^2)/((3t^2+1)(t^2+1))] dt
use int by parts fraction
(2-2t^2)/((3t^2+1)(t^2+1)) = A/(3t^2+1) + B/(t^2+1)
gives A=4, B=-2
continue it ...... :p

Answer 8

ohh am so sorry! damn!!