Question

How do I evaluate `int(1-sinx)/cosx dx`?

Answer 1

`=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))`; so I get:

`int(1-sin(x))/cos(x)*(1+sin(x))/(1+sin(x))dx=`
`=int(1-sin^2(x))/(cos(x)(1+sin(x)))dx=`
`=int(cos^2(x))/(cos(x)(1+sin(x)))dx=`
`=int(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`

Which derived gives (1) or your original function!!!!!