Question

How do I evaluate `int(1 + cosx)/sinx d x`?

Answer 1

`=-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))`; so I get:

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

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