- 1,062

- 6

Show that the integral of

__sin(θ)-cos(θ)__dθ

sin(θ)+cos(θ)

= -(1/2)ln(2sin(2θ)+2)+C

Now what I did was let u=sin(θ)+cos(θ) so that du=cos(θ)-sin(θ)dθ or that du=-(sin(θ)-cos(θ))dθ

Now I got the integral of

__(sin(θ)-cos(θ))*du__

u*-(sin(θ)-cos(θ))

And then both of the sin(θ)-cos(θ) cancels out and I was left with the integral of

__-du__

u

which is -ln|u|+C , plugging back what u was into that I got

-ln|sin(θ)+cos(θ)|+C

which I cannot figure out to be equal to what he said it should be.

The only thing I can think of is the C's being different and that causing them to be equal. Any ideas?