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
