# Graphical integration of $$\sin^2(x)$$ and $$\cos^2(x)$$

## 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 $$\sin^2(x)+\cos^2(x)=1$$.