I don't think that #cos(x^6)# has an antiderivative expressible with elementary functions, so I assume you want to find #int\ cos^6(x)\ dx#.
First, recall that #cos(2theta)=2cos^2(theta)-1#, or #cos^2(theta)=(1+cos(2theta))/2#.
So, we have
#\ \ \ \ \ \ int\ cos^6(x)\ dx#
#=int\ (cos^2(x))^3\ dx#
#=int\ ((1+cos(2x))/2)^3\ dx#
#=1/8int\ (1+cos(2x))^3\ dx#
#=1/8int\ (1+3cos(2x)+3cos^2(2x)+cos^3(2x))\ dx#
#=1/8(int\ dx+3int\ cos(2x)\ dx+3int\ cos^2(2x)\ dx+int\ cos^3(2x))\ dx)#
Let's look at the integrals one by one.
The first integral is obvious:
#\ \ \ \ \ \ int\ dx=x+C#
The second integral requires the reverse chain rule:
#\ \ \ \ \ \ 3int\ cos(2x)\ dx#
#=3int\ cos(2x)dx/(d(2x))\ d(2x)#
#=3/2int\ cos(2x)\ d(2x)#
#=3/2sin(2x)+C#
The third integral requires using #cos^2(theta)=(1+cos(2theta))/2# again:
#\ \ \ \ \ \ 3int\ cos^2(2x)\ dx#
#=3int\ (1+cos(4x))/2\ dx#
#=3/2int\ (1+cos(4x))\ dx#
#=3/2x+3/2int\ cos(4x)\ dx#
#=3/2x+3/2int\ cos(4x)dx/(d(4x))\ d(4x)#
#=3/2x+3/8int\ cos(4x)\ d(4x)#
#=3/2x+3/8sin(4x)+C#
The fourth integral requires the knowledge that #sin^2(theta)+cos^2(theta)=1# and the reverse chain rule again:
#\ \ \ \ \ \ int\ cos^3(2x)\ dx#
#=int\ cos(2x)(1-sin^2(2x))\ dx#
#=int\ cos(2x)(1-sin^2(2x))dx/(d(sin(2x)))\ d(sin(2x))#
#=1/2int\ (1-sin^2(2x))\ d(sin(2x))#
#=1/2sin(2x)-1/6sin^3(2x)+C#
Put all of them together to get
#5/16x+1/4sin(2x)+3/64sin(4x)-1/48sin^3(2x)+C#
