How do you find the antiderivative of #int sinx(cosx)^(3/2) dx#?

1 Answer
Jan 14, 2017

#-2/5(cosx)^(5/2)+C#

Explanation:

Since we don't want to have to deal with the #3//2# power, let #u=cosx#. This implies that #du=-sinxcolor(white).dx#.

So, we need to modify our integral just a little:

#intsinx(cosx)^(3/2)dx=-int(cosx)^(3/2)(-sinxcolor(white).dx)=-intu^(3/2)color(white).du#

Integrate this using the rule #intu^ncolor(white).du=u^(n+1)/(n+1)#, where #n!=-1#.

#=-u^(5/2)/(5/2)=-2/5u^(5/2)=-2/5(cosx)^(5/2)+C#