how do i solve the integral of sin(ax) cos(ax)?
i don't understand the method of goniometric integrals
Stacey Warren - Expert brainly.com
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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...
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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.
nice, you made me get it, I hope i can put this to practice, thanks!
No problem; I'm glad I could help. :)
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
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 .
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.