=ln(1+sin(x))+c=ln(1+sin(x))+c
Well, this one is 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+sin(x))/(1+sin(x))1+sin(x)1+sin(x); so I get:
int(1-sin(x))/cos(x)*(1+sin(x))/(1+sin(x))dx=∫1−sin(x)cos(x)⋅1+sin(x)1+sin(x)dx=
=int(1-sin^2(x))/(cos(x)(1+sin(x)))dx==∫1−sin2(x)cos(x)(1+sin(x))dx=
=int(cos^2(x))/(cos(x)(1+sin(x)))dx==∫cos2(x)cos(x)(1+sin(x))dx=
=int(cos(x))/((1+sin(x)))dx==∫cos(x)(1+sin(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+sin(x))+c=ln(1+sin(x))+c
Which derived gives (1) or your original function!!!!!
