# Thread: reduction formula (application of integration by parts)

1. ## reduction formula (application of integration by parts)

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.

2. 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.

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 $\int sin^2x cos^4 x dx= uv- \int v du= -(1/4)cos^4 x sin x- (1/4)cos^5 x dx$.