#intcos^5(x)dx#
If you know the complex exponential form of #cos(x)# then you could proceed using the following method:
#cos^5(x)=((e^(ix)+e^(-ix))/(2))^5#
Now, expand this with the binomial theorem to get:
#=1/32(e^(i5x)+5e^(i3x)+10e^(ix)+10e^(-ix)+5e^(-3ix)+e^(-5ix))#
Now collect the terms like so:
#=1/32({e^(i5x)+e^(-i5x)}+5{e^(i3x)+e^(-i3x)}+10{e^(-ix)+e^(ix)})#
#=1/16({e^(i5x)+e^(-i5x)}/2+5{e^(i3x)+e^(-i3x)}/2+10{e^(-ix)+e^(ix)}/2)#
#=1/16cos(5x)+5/16cos(3x)+5/8cos(x)#
So from this it may follow that:
#intcos^5(x)dx=int1/16cos(5x)+5/16cos(3x)+5/8cos(x)dx#
Now integrate to get:
#1/80sin(5x)+5/48sin(3x)+5/8sin(x)+C#