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

1 Answer
Oct 20, 2015

I = (6sin^7x)/7-(2sin^9x)/3+C

Explanation:

I = int (6sin^6x)(cos^3x)dx = 6int sin^6x cos^2x cosxdx

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

sinx=t => cosxdx=dt

I = 6int t^6 (1-t^2) dt = 6 int (t^6-t^8)dt = 6 (t^7/7-t^9/9+C)

I = (6sin^7x)/7-(2sin^9x)/3+C