Question

How do you evaluate the integral `∫ (1-sinx)/cosx dx`?

Answer 1

Well, this one is tough and I'll need to use a trick.

I multiply and divide by `1+sin(x)`!!!
So you get:

`int[(1-sin(x))(1+sin(x))]/[cos(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=`

but `d[sin(x)]=cos(x)dx`

and:
`=int[1]/[1+sin(x)]d[sin(x)]=` and finally:

`=ln|1+sin(x)|+c`

Hope it helps