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# Integrate sinxsin2x dx watch

1. I get 1/2sinx - 1/6sin3x

Anyone confirm?
2. (Original post by imasillynarb)
I get 1/2sinx - 1/6sin3x

Anyone confirm?
I get 2/3 * sin^3(x) + k
3. int{sinx.sin(2x)} dx

sin(2x) = 2sinx.cosx

2.int{sin^2(x).cosx}dx

let u = sinx
du/dx = cosx

2.int{u^2}du

2(u^3)/3 + C

(2/3).sin^3(x) + C

I might have made a mistake but our constants could just be different. :-/
4. I did 2sinAsinB = cos(A-B) - cos(A+B)

sin2xsinx = (cosx - cos3x)/2

Integrate to get 1/2sinx - 1/6sin3x

Cant see anything wrong with that??
5. (Original post by imasillynarb)
I did 2sinAsinB = cos(A-B) - cos(A+B)

sin2xsinx = (cosx - cos3x)/2

Integrate to get 1/2sinx - 1/6sin3x

Cant see anything wrong with that??
6. (Original post by Sahir)
I thought as much, although I couldnt figure out how to rearrange it and show it
7. (Original post by imasillynarb)
I thought as much, although I couldnt figure out how to rearrange it and show it
Heh yeah me neither.. i sub'd in a few random numbers for x, shoved it into a calculator, and tada!
8. METHOD 1
(int) sin(x) sin(2x) dx
= (int) 2sin^2(x) cos(x) dx
= (2/3) sin^3(x) + constant.

METHOD 2
sin(A) sin(B) = (1/2) [ cos(A - B) - cos(A + B) ]

(int) sin(x) sin(2x) dx
= (int) (1/2) [ cos(x) - cos(3x)] dx
= (1/2) sin(x) - (1/6) sin(3x) + constant.

CHECK THE TWO ANSWERS ARE THE SAME
cos(3x) + i sin(3x) = (cos(x) + i sin(x))^3

Taking the imaginary part,

sin(3x)
= 3cos^2(x) sin(x) - sin^3(x)
= 3sin(x) - 4sin^3(x).

So (1/2) sin(x) - (1/6) sin(3x) = (2/3) sin^3(x).

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