- #1
mattmns
- 1,128
- 6
Ok here it is,
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?
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?