Question

how do i solve the integral of sin(ax) cos(ax)?
i don't understand the method of goniometric integrals

Answer 1

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...

Answer 2

why do you do du=a dx?

Answer 3

Just to get the 1/a in front.

Answer 4

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.

Answer 5

nice, you made me get it, I hope i can put this to practice, thanks!

Answer 6

No problem; I'm glad I could help. :)

Answer 7

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

Answer 8

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 .

Answer 9

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.