# Integral Math problem

mattmns
Ok here it is,

Show that the integral of

sin(&theta;)-cos(&theta;)d&theta;
sin(&theta;)+cos(&theta;)

= -(1/2)ln(2sin(2&theta;)+2)+C

Now what I did was let u=sin(&theta;)+cos(&theta;) so that du=cos(&theta;)-sin(&theta;)d&theta; or that du=-(sin(&theta;)-cos(&theta;))d&theta;

Now I got the integral of

(sin(&theta;)-cos(&theta;))*du
u*-(sin(&theta;)-cos(&theta;))

And then both of the sin(&theta;)-cos(&theta;) 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(&theta;)+cos(&theta;)|+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?

## Answers and Replies

Staff Emeritus
Gold Member
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(2&theta;)+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(2&theta;)+2)+C?

Last edited:
Staff Emeritus