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Thread: reduction formula (application of integration by parts)

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    Talking reduction formula (application of integration by parts)

    By finding a reduction formula for $\displaystyle sin^2x cos^4 x $, evaluate the integral.

    can someone show me how to do this? I don't need help with the evaluation, but I don't know how to find the reduction formula.

    Full working out, please

    Thanks in advance
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    Quote Originally Posted by differentiate View Post
    By finding a reduction formula for $\displaystyle sin^2x cos^4 x $, evaluate the integral.

    can someone show me how to do this? I don't need help with the evaluation, but I don't know how to find the reduction formula.

    Full working out, please

    Thanks in advance
    There are actually two ways to proceed. Since (sin x)'= cos x, you can take $\displaystyle u= cos^3 x$ so that $\displaystyle du= 3 cos(x)dx$ and $\displaystyle dv= sin^2 x cos x dx$ so that, with y= sin x, $\displaystyle dv= u^2 du$ and $\displaystyle v= (1/3)u^3= (1/3)sin^3 x dx$
    Then $\displaystyle \int sin^2x cos^4 x dx= uv- \int v du= (1/3)cos^3 x sin^3 x- \int sin^3 x cos(x)dx$

    Or, since (cos x)'= -sin x, you can take $\displaystyle u= sin x$ so that $\displaystyle du= cos x dx$ and $\displaystyle dv= cos^3 x sin x dx$ so that $\displaystyle v= -(1/4) cos^4x$.
    Then $\displaystyle \int sin^2x cos^4 x dx= uv- \int v du= -(1/4)cos^4 x sin x- (1/4)cos^5 x dx$.
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