How do you find the integral of #int cos^3(x)*sin^6(x) dx#?

1 Answer
Oct 11, 2015

This belongs in your mathematical toolbox: #int sin^m u cos^n u du# with at least one of #m, n# odd.

Explanation:

#int sin^m u cos^n u du# with at least one of #m, n# odd.
Integrate by substitution. Do this by pulling off one from the odd power, then convert the remaining even power to the other function. Integrate the resulting polynomial in #sinu# or #cosu# term by term.

#I = int cos^3xsin^6x dx = int cos^2xsin^6x cosxdx#

# = int (1-sin^2x)sin^6x cosxdx#

# = int (sin^6x-sin^8x) cosxdx#

# = sin^7x/7 - sin^9/9x +C#