Calc II - how to evaluate the integral of sin^3(2x)dx?

Question

Hi,

I need to evaluate the integral of sin^3(2x)dx.

I started by using the power-reducing formula to turn sin^2 x into (1-cos 4x/2) and multiplying that by sin 2xdx.

So I have: ⌡ (1-cos 4x/2) (sin 2x) dx

Then I took the 1/2 out front to get:
1/2 ⌡ (1-cos 4x) (sin 2x) dx

Then I distributed the sin 2x out to get:
1/2 ⌡ (sin 2x - sin 2xcos4x) dx

Well, I wanted to use u-sub. with u = sin 2x and du = 2 cos 2x, but as you can see, I have cos 4x. How would I complete this problem (if I was on the right track in the first place)?

Thanks for the help! :>)

Answer 1

It's easier if you write

sin^3(2x) = sin(2x) sin^2(2x)
= sin(2x) (1 - cos^2(2x))
= sin(2x) - sin(2x) cos^2(2x)

Then you are left with elementary integrals and you get:

-1/2 cos(2x) + 1/6 cos^3(2x)

Answer 2

Dérivés de sin^3(2x)

Answer 3

Try this:

sin^3(2x) =sin^3(u) = (1-cos^2(u))sin(u)

du = 2dx

Integral(sin^3(2x)dx)=integral([1-cos^2(u)]sin(u)du/2)

Let v = cos(u) ---> dv =-sin(u)du

I = Integral([1-cos^2(u)]sin(u)du/2) = integral([1-v^2](-dv/2))

I = -v/2 +v^3/6 =-cos(2x)/2 +cos^3(2x)/6

Answer 4

first writing sin^3(2x) as (1-cos*2 2x) sin2x
u=cos2x ,-1/2du=sin2xdx
integral of -1/2(1-u*2)du

after solving u will get--
(cos2x)*3 /6 -(cos2x)/2
hope this will help u...........

Answer 5

this is the formula:
⌡sin^3(ax)dx. = 1/3a cos(ax) { sin^2(ax)+2}

⌡sin^3(2x)dx. = 1/6 cos(2x) { sin^2(2x)+2} ............

Hope this would help...all the best

Answer 6

use trigonometric transformation:
sen^3(x)=3/4·sin(x)-1/4·sin(3x)
replace x ->>2x
sen^3(2x)=3/4·sin(2x)-1/4·sin(6x)

int(sen^3(2x))=int(3/4·sin(2x)-1/4·sin(6x))
__________=-3/8·cos(2x)+1/24·cos(6x)

Good luck!