reduction formula (application of integration by parts)

**differentiate** · April 3rd 2010, 04:26 AM

By finding a reduction formula for $sin^2x cos^4 x$ , 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

**HallsofIvy** · April 3rd 2010, 04:44 AM

Originally Posted by differentiate

By finding a reduction formula for $sin^2x cos^4 x$ , 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 $u= cos^3 x$ so that $du= 3 cos(x)dx$ and $dv= sin^2 x cos x dx$ so that, with y= sin x, $dv= u^2 du$ and $v= (1/3)u^3= (1/3)sin^3 x dx$
Then $\int sin^2x cos^4 x dx= uv- \int v du= (1/3)cos^3 x sin^3 x- \int sin^3 x cos(x)dx$

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

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