How do you find the antiderivative of #sin^3(x) cos^5(x) dx#?
1 Answer
Nov 16, 2016
# int sin^3xcos^5xdx = 1/4sin^4x-1/3sin^6x+1/8sin^8x + C#
Explanation:
Let
Using the trig identity
So we can substitute into
# int sin^3xcos^5xdx = int sin^3x(cos^2x)^2cosxdx #
# :. int sin^3xcos^5xdx = int u^3(1-u^2)^2du #
# :. int sin^3xcos^5xdx = int u^3(1-2u^2+u^4)du #
# :. int sin^3xcos^5xdx = int u^3-2u^5+u^7du #
# :. int sin^3xcos^5xdx = 1/4u^4-1/3u^6+1/8u^8 + C#
# :. int sin^3xcos^5xdx = 1/4(sinx)^4-1/3(sinx)^6+1/8(sinx)^8 + C#
# :. int sin^3xcos^5xdx = 1/4sin^4x-1/3sin^6x+1/8sin^8x + C#
Substituting any other trig function would also yield the same result.