Thread: reduction formula (application of integration by parts)

1. 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.

2. Originally Posted by differentiate
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.

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