Proving Integral of sin(θ) - cos(θ) = -(1/2)ln(2sin(2θ)+2)+C

[mattmns]

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?

 

[Hurkyl]

Notice that their solution has a (1/2) out front. Try multiplying your solution by 2/2. (since 2/2 = 1, this doesn't change the solution)

Now, leave the (1/2) outside the logarithm, but use the power rule (n log a = log (a^n)) to pull the 2 inside the logarithm, and see if you can manipulate the result to get their answer. You'll probably be off by a multiplicative constant in the logarithm, but then you can use the fact that log ab = log a + log b to pull it out and absorb it into C.

 

[mattmns]

Of course mutliply by one. I always forget to multiply by 1, but I never forget to add 0. Thanks

Edit...

Ok I now got to this answer

-(1/2)ln(sin(2θ)+1)+C

Now can I add -(1/2)ln(2)? And then pull the 2 inside the original log and make a new constant C? to get the answer I wanted of -(1/2)ln(2sin(2θ)+2)+C?

 

[Hurkyl]

Bingo!