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    EDIT: Got it now, it's splitting sin^3(x) dx into sinx(sin^2x) dx and as du=-sinx dx it changes to -(1-cos^2x)du = -(1-u^2)du.

    The question is: find the integral of sin^3(x) dx, using u=cosx as a substitution.

    They've given this as the working out method and final answer.




    I don't quite understand it, it's replacing sin^3(x) with (1-u^2) meaning sin^3(x) = 1 - cos^2(x)....which isn't quite right, as this is the trig identity for sin^2(x) not sin^3(x)


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    sin^3(x)=sin(x)sin^2(x)
    sin^2(x) = 1-cos^2(x)
    sin(x)dx=-du
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    (Original post by Panda Vinnie)
    EDIT: Got it now, it's splitting sin^3(x) dx into sinx(sin^2x) dx and as du=-sinx dx it changes to -(1-cos^2x)du = -(1-u^2)du.

    The question is: find the integral of sin^3(x) dx, using u=cosx as a substitution.

    They've given this as the working out method and final answer.




    I don't quite understand it, it's replacing sin^3(x) with (1-u^2)
    No,no. You replace sin^2(x) with (1-u^2)
    THe 'third' sinx together with the dx integrating factor is in the -du,
    as it is sinx*dx
 
 
 
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