Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 14

intcos(theta)cos^5(sin(theta))d theta
Apply integral substitution,
Let u=sin(theta)
=>du=cos(theta)d theta
=intcos^5(u)du
Rewrite the above integrand as,
=intcos^4(u)cos(u)du
Now use the identity: cos^2(x)=1-sin^2(x)
=int(1-sin^2(u))^2cos(u)du
Apply the integral substitution,
Let v=sin(u)
=>dv=cos(u)du
=int(1-v^2)^2dv
=int(1-2v^2+v^4)dv
=int1dv-2intv^2dv+intv^4dv
=v-2(v^3/3)+v^5/5
Substitute back v=sin(u)
=sin(u)-2/3sin^3(u)+1/5sin^5(u)
Substitute back u=sin(theta)
=sin(sin(theta))-2/3sin^3(sin(theta))+1/5sin^5(sin(theta))
Add a constant C to the solution,
=sin(sin(theta))-2/3sin^3(sin(theta))+1/5sin^5(sin(theta))+C