How to calculate average of sine squared

I have problem with calculating average of sine. In my book, there is claim, that $$\langle\sin^2\omega t\rangle=\frac12$$ Now I have a question how to get this result mathematically? I know you must integrate by cycle, but I can not get to this result

Solution 1:

There is a nice trick. You know that:

$$\sin^2 \omega t+\cos^2 \omega t = 1$$

Calculate the average of this equalty, since the average over a cycle is the same for the sine and the cosine and $\langle 1 \rangle = 1$:

$$\langle \sin^2 \omega t \rangle= \frac 1 2$$

Solution 2:

The mean of a function over an interval $T$ is given by,

$$\langle f(t) \rangle = \frac{1}{T}\int_0^T f(t) \, \mathrm{d}t$$

which is roughly the continuous analogue of the arithmetical mean. For $\sin^2 \omega t$, we obtain,

$$\langle \sin^2 \omega t \rangle = \frac{1}{T} \int_0^T \sin^2 \omega t \, \mathrm{d} t = \frac{1}{T} \left( \frac{T}{2} - \frac{\sin(2\omega T)}{4\omega}\right)$$

If we choose $T=2\pi \omega^{-1}$, the result simplifies to,

$$\langle \sin^2 \omega t \rangle = \frac{1}{2}.$$

Solution 3:

\begin{align} &\frac{1}{T}\int_{-T/2}^{+T/2}\sin^2\omega t\,dt\\ &\overset{1-2\sin^2\omega t=\cos2\omega t}{=} \frac{1}{T}\int_{-T/2}^{+T/2}\left(\frac{1}{2}-\frac{1}{2}\cos2\omega t\right)\,dt\\ &=\frac{1}{T}\left(\frac{1}{2}t\big|_{-T/2}^{T/2}-\frac{1}{4\omega}\sin2\omega t\big|_{-T/2}^{+T/2}\right)\\ &\overset{T\to+\infty,\,\omega=\frac{2\pi}{T}}{=}\frac{1}2-\frac{1}{T}\cdot\frac{T}{8\pi}\sin\left(\frac{4\pi}{T}\cdot\frac{T}{2}\right)\cdot2\\ &=\frac{1}{2} \end{align}