Visualisation; beats using double angle formulas!
Here's a nice way of visualising the integration of \(\sin^2(x)\) and \(\cos^2(x)\) between \(\pi\) and \(2\pi\). Below the graphic is the algebraic version, which involves the use of double angle formulas. I don't like to condone "tricks" over understanding, but this is quite a nice demonstration I saw not long ago for speed of remembering!
...and now the slightly more gritty written solution
First use the double angle formula for \(\cos(2x)\) to express \(\sin^2(x)\) differently: \[\begin{align} \cos(2x)&=\cos^2(x)-\sin^2(x) \\[8pt] &=(1-\sin^2(x))-\sin^2(x) \\[8pt] &=1-2\sin^2(x) \\[8pt] 2\sin^2(x)&=1-\cos(2x) \\[8pt] \sin^2(x)&=\tfrac{1}{2}-\tfrac{1}{2}\cos(2x) \end{align}\] Then integrate using the substitution from above: \[\begin{align} &\int^{2\pi}_0 \sin^2(x)\,dx \\[12pt] &= \int^{2\pi}_0 \tfrac{1}{2}-\tfrac{1}{2}\cos(2x) \,dx \\[12pt] &= \left[\tfrac{1}{2}x-\tfrac{1}{4}\sin(2x)\right]^{2\pi}_0 \\[12pt] &= \left(\tfrac{1}{2}(2\pi)-\tfrac{1}{4}\sin(4\pi)\right)-\left(\tfrac{1}{2}(0)-\tfrac{1}{4}\sin(0)\right) \\[12pt] &= \pi \end{align}\]
Intergrating \(\cos^2(x)\) is equivalent, but you will need to substitute out the \(\sin^2(x)\) term instead on the second line. In case you didn't realise, this uses \(\sin^2(x)+\cos^2(x)=1\).
