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Integral Math problem

  • Thread starter mattmns
  • Start date
1,058
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?
 

Answers and Replies

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,843
17
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.
 
1,058
6
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?
 
Last edited:
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,843
17
Bingo!
 

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