`int cos(theta)cos^5(sin(theta)) d theta` Evaluate the integral

`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`