How do you integrate #sin^2(x) cos^5(x)#?

1 Answer
Jim H
Oct 29, 2015

Recommended: learn this pattern.

Explanation:

To evaluate #int sin^mxcos^nx dx# where #m,n# are positive integers and at least one of #m,n# is odd.

Pull off one from the odd power. (If both are odd, it is simpler, but not necessary, to make the lesser power even.)
Change the remaining even power to the other function using #sin^2x+cos^2x = 1#.

Expand and substitute to get a polynomial in #u#

It may sound complicated, but here's how it works for this question:

#intsin^2xcos^5xdx=intsin^2xcos^4xunderbrace(cosxdx)_(du)#

We are going to use #u = sinx# so we need to rewrite

#cos^4x = (cos^2x)^2 = (1-sin^2x)^2 = 1-2sinx+sin^2x#

Our integral becomes:

#intsin^2xunderbrace(( 1-2sinx+sin^2x))_(cos^4x) cosx dx#

# = int (sin^2x-2sin^3x+sin^4x) cosx dx#

# = int (u^2-2u^3+u^4) du = u^3/3+u^4/2+u^5/5+C#

# = sin^3x/3+sin^4x/2+sin^5x/5 +C#

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