=-ln(1+cos(x))+c=−ln(1+cos(x))+c
Well, this one is another one a little bit tricky...
I started with the idea that the result must be a logarithm...
So I did a little manipulation to get to a friendlier version of your function by multiplying and dividing by:
(1+cos(x))/(1+cos(x))1+cos(x)1+cos(x); so I get:
int(1-cos(x))/sin(x)*(1+cos(x))/(1+cos(x))dx=∫1−cos(x)sin(x)⋅1+cos(x)1+cos(x)dx=
=int(1-cos^2(x))/(sin(x)(1+cos(x)))dx==∫1−cos2(x)sin(x)(1+cos(x))dx=
=int(sin^2(x))/(sin(x)(1+sin(x)))dx==∫sin2(x)sin(x)(1+sin(x))dx=
=int(sin(x))/((1+cos(x)))dx==∫sin(x)(1+cos(x))dx= which is a manipulated version of your original function and we can call it (1). The integral of (1) is indeed a logarithm:
=-ln(1+cos(x))+c=−ln(1+cos(x))+c
Which derived gives (1) or your original function!!!!!
