That seems abit longer and harder than msjds method.
His way easily goes to this:
[tex]\frac{1}{2} \int 1 dx + \int \cos 2x dx = \frac{1}{2} (x + \int cos 2x dx)[/tex]
u=2x du/dx = 2
[tex]\frac{1}{2}(x+\frac{1}{2}\int cos u du)= \frac{1}{2} ( x+\frac{1}{2}\sin u)[/tex]We're done, pretty fast too :)
[tex]\int \cos^2 x dx = \frac{x}{2} + \frac{\sin 2x}{4}[/tex]
BTW emilgouliev, nice work, and Welcome to Physicsforums. :)