Question

What's the integral of `int tanx / (secx + cosx)dx`?

Answer 1

I found: `-arctan[cos(x)]+c`

Explanation:

We can try changing everything into sin and cos to get:
`int(sin(x)/cos(x))/(1/cos(x)+cos(x))dx=`
`=int(sin(x)/cancel(cos(x)))*(cancel(cos(x))/(1+cos^2(x)))dx=`
`=intsin(x)/(1+cos^2(x))dx=`
consider that `d[cos(x)]=-sin(x)dx` so we have:
`=-int1/(1+cos^2(x))d[cos(x)]=` with `cos(x)` as "variable" of integration to get;
`=-arctan[cos(x)]+c`