anonymous
  • anonymous
how do i solve the integral of sin(ax) cos(ax)? i don't understand the method of goniometric integrals
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
First, let u=ax and du=a dx. dx = du/a, so you can simplify it to being 1/a * ∫ sin(u)cos(u) du. Now, I'm basically going to do another u-sub, but for clarity going to call it v: v = cos(u), dv = -sin(u) du. Now, du = -dv/sin(u), and when you plug the very last phrase into your integral, you have -1/a * ∫v dv. Then, solve and re-substitute: -1/a * v^2/2 + c = -1/a * cos(u)^2/2 + c = -1/(2a) * cos(ax)^2 + c. Not really sure what you mean by "goniometric" though...
anonymous
  • anonymous
why do you do du=a dx?
anonymous
  • anonymous
Just to get the 1/a in front.

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anonymous
  • anonymous
I don't think it's absolutely necessary, but it makes things a little cleaner during your second substitution. The absolutely crucial sub here is v = cos(u), dv = -sin(u) du because that eliminates sin(u) from your integral.
anonymous
  • anonymous
nice, you made me get it, I hope i can put this to practice, thanks!
anonymous
  • anonymous
No problem; I'm glad I could help. :)
anonymous
  • anonymous
what i meant with goniometric, my study method does something else: F sin ax cos ax dx = 1/a F cos ax d(-cos ax) = -1/a F cos ax d cos ax = -1/2a cos^2 ax +C i understand your method, just not the one we are supposed to know
anonymous
  • anonymous
an easy method is u=ax then:1/a * ∫ sin(u)cos(u) du so sin(u)cos(u)=(1/2)*sin(2u) and then the integral equals (1/2a )* ∫ sin(2u)du=(-1/4a)*cos(2u)+c .
anonymous
  • anonymous
ijadi, I'm not sure why but according to Wolfram|Alpha and another computerized integrator our answers are not congruent, even though I can't see anything wrong with your method. Strange. Meester, the only thing I meant was that I've never heard of goniometric before in my life! xD I don't see exactly why that name is attached but I understand how you did that integral now.

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