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I hate math help pls........how do i solve the integral from 0-pi/4 of sin(2x)sin(x)dx?

I tried to work this out by setting u = to sin(x) then du was = to cos(x).........my final answer was sqrt(2)/3 but this was not right......

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  • 1 decade ago
    Favorite Answer

    following from the person above me,

    set u=cosx then du= -sinx

    then u^2(-du)

    = -1/3u^3

    do you mean 0 to pi/4

    then -1/3(1/sqrt(2))^(3)= -1/3[2sqrt(2)]-1/3

    = [-1-2sqrt(2)]/[6sqrt(2)]

  • Anonymous
    1 decade ago

    Double angle formula.....

    sin(2x)=2sin(x)cos(x)

    so therefore...integral sin^2(x)cos(x).....I'm sure you can take it from there

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