- $\displaystyle \int(sin^2(x))^2 cos^2(x) dx$
- $\displaystyle \int(\frac{1-cos(2x)}{2})^2(\frac{1+cos(2x)}{2}dx$
- $\displaystyle \frac{1}{8}\int1-cos(2x)-cos^2(2x)+cos^3(2x)dx$
- $\displaystyle \frac{1}{8}\int1dx-\frac{1}{8}\int cos(2x)dx-\frac{1}{8}\int cos^2(2x)dx+\frac{1}{8}\int cos^3(2x)dx$
- $\displaystyle \frac{1}{16}x-\frac{1}{64}sin(4x)-\frac{1}{48}sin^3(2x)+C$

What I dont understand is how he got from 2 to 3 and from 4 to 5. For three, I had this:

$\displaystyle \frac{1}{8}\int-c^3-2c^2+c+1dx$ Where c is a place holder for cos 2x.