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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?

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    $\begingroup$ 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. $\endgroup$ – vrugtehagel May 8 at 16:58
  • $\begingroup$ 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. $\endgroup$ – Michael Seifert May 8 at 17:04
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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 $$ \int_{0}^{2\pi} \sin^{2} t dt = \int_{0}^{2\pi} \cos^{2}\left(t-\frac{\pi}{2}\right) \,dt = \int_{-\frac{\pi}{2}}^{2\pi - \frac{\pi}{2}} \cos^{2}t\,dt = \int_{0}^{2\pi} \cos^{2}t\,dt $$

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

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