# Show $\int_0^{2\pi}\cos^2t\,dt=\pi$ using Pythagorean Theorem

The 'traditional' way I always see this integral calculated is with the identity

$$\cos^2t=\frac{1+\cos(2t)}{2}$$

My alternative method uses $$\cos^2t+\sin^2t=1$$. It's obvious that

$$\int_0^{2\pi}\cos^2t+\sin^2t\,dt=\int_0^{2\pi}1\,dt=2\pi$$

The interval of integration is an integer multiple of the periods of each function ($$\cos^2t$$ has a period of $$\pi$$), and so it seems reasonable to me that given the Pythagorean identity used above, $$\cos^2t$$ and $$\sin^2t$$ contribute, for lack of a better word, equally to this final answer of $$2\pi$$ above, and so the integral in the title should be half of $$2\pi$$, or just $$\pi$$.

Is this method valid? What additional statements, if any, are necessary to make it rigorous enough to be valid?

• One way to make it more rigorous is just to shift the integral for $\sin^2$ by $\pi/2$ so that you just get the integral for $\cos^2$ twice. – vrugtehagel May 8 at 16:58
• One of my high-school teachers showed me this trick for calculating the average of $\cos^2 x$, and it's stuck with me ever since. – Michael Seifert May 8 at 17:04

## 2 Answers

Your argument is definitely valid. To add more explanation, we can say that $$\int_{a}^{a+T} f(x)dx = \int_{0}^{T} f(x) dx$$ for any $$T$$-periodic function $$f(x)$$ (try to prove this rigorously), and then $$\int_{0}^{2\pi} \sin^{2} t dt = \int_{0}^{2\pi} \cos^{2}\left(t-\frac{\pi}{2}\right) \,dt = \int_{-\frac{\pi}{2}}^{2\pi - \frac{\pi}{2}} \cos^{2}t\,dt = \int_{0}^{2\pi} \cos^{2}t\,dt$$

Use $$\displaystyle\int_0^{2a}f(x)\ dx=2\int_0^af(x) \ dx$$ for $$f(2a-x)=f(x)$$

twice to find $$I=\int_0^{2\pi}\cos^2t\ dt=4\int_0^{\pi/2}\cos^2t\ dt$$

Now $$\displaystyle\int_a^bf(x) \ dx=\int_a^bf(a+b-x) \ dx$$

to find $$2\cdot\dfrac I4=\int_0^{\pi/2}(\cos^2t+\sin^2t)dt$$