How do you evaluate the integral #int sin^3xsqrt(1-cosx)#?

Question

1314 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-20T10:29:13

#int sin^3x sqrt (1-cosx)dx = 2/7(1-cosx)^(5/2) (9/5+cosx) + C #

Write:

#sin^3x = sinx * sin^2x = sinx (1-cos^2x) = sinx (1-cosx)(1+cosx)#

so that:

#int sin^3x sqrt (1-cosx)dx = int sinx (1+cosx)(1-cosx)^(3/2) dx#

substitute:

#t = 1-cosx#
#dt = sinx dx#

#int sin^3x sqrt (1-cosx)dx = int t^(3/2)(2-t)dt= 2int t^(3/2)dt - int t^(5/2)dt = 4/5t^(5/2) - 2/7t^(7/2) + C = 2/7t^(5/2) (14/5-t) + C#

Substituting back:

#int sin^3x sqrt (1-cosx)dx = 2/7(1-cosx)^(5/2) (9/5+cosx) + C #