How do you evaluate the integral of #x*(cos(x^2))^5#?

First, let

#u=x^2" "=>" "du=2xdx#

So, we have

#intxcos^5(x^2)dx#

Rearrange the #x# and #dx# and add a #2# to achieve #2xdx#. To balance the #2# inside the integral, multiply by #1//2# outside of the integral.

#=1/2intunderbrace(cos^5(x^2))_(cos^5(u))*underbrace(2xdx)_(du)#

Substituting, this equals

#=1/2intcos^5(u)du#

Before we integrate this, we want to modify this so that we can have a #v# and #dv# term both in the integral. We can do this by writing #cos^5(u)# as

#cos^5(u)=cos(u)cos^4(u)=cos(u)(cos^2(u))^2=cos(u)(1-sin^2(u))^2#

Giving us the integral

#=1/2intcos(u)(1-sin^2(u))^2du#

Now, we can substitute again:

#v=sin(u)" "=>" "dv=cos(u)du#

Rearrange the integral:

#=1/2intunderbrace((1-sin^2(u))^2) _ ((1-v^2)^2) * underbrace(cos(u)du) _ (dv)#

Which equals:

#=1/2int(1-v^2)^2dv#

Distribute #(1-v^2)^2#.

#=1/2int(1-2v^2+v^4)dv#

Splitting up the integral, we obtain:

#=1/2intv^4dv-2(1/2)intv^2dv+1/2int1dv#

Using the rule:

#intv^ndv=v^(n+1)/(n+1)+C#

Yields, after integrating term by term:

#=1/2(v^5/5)-v^3/3+1/2(v)+C#

#=v^5/10-v^3/3+v/2+C#

Using #v=sin(u)#:

#=sin^5(u)/10-sin^3(u)/3+sin(u)/2+C#

Now using #u=x^2#:

#=sin^5(x^2)/10-sin^3(x^2)/3+sin(x^2)/2+C#