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    I get 1/2sinx - 1/6sin3x

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

    Anyone confirm?
    I get 2/3 * sin^3(x) + k
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    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. :-/
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    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??
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    (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??
    Yeah your answer is identical to (2/3)(sin x)^3
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    (Original post by Sahir)
    Yeah your answer is identical to (2/3)(sin x)^3
    I thought as much, although I couldnt figure out how to rearrange it and show it
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    (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!
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    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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