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
 A: \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}
A: Another way to see this is by using the fact that $$\sin^2 \omega t=\frac{1-\cos(2\omega t)}{2}$$
and the average value of $\cos(2\omega t)$ is zero.
A: The average of discrete quantities is:
$$\langle x \rangle = \frac{1}{N}\sum_i x_i$$
The extension to a continuous quantity should seem fairly natural if you think of an integral as a continuous sum:
$$ \langle f(x) \rangle = \frac{1}{b-a} \int_a^b f(x) dx$$
where $[a,b]$ is the interval you want to average over. Then you just need to evaluate this integral. The bounds $(-\infty,\infty)$ make it seem a bit difficult, but for a periodic function the average over one cycle is the same as the average over many cycles.
A: 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 $\langle 1 \rangle = 1$:
$$\langle \sin^2 \omega t \rangle= \frac 1 2$$ 
A: It can be easily answered graphically by realizing $\displaystyle \sin^2(x) - \frac{1}{2} $ is symmetric about the x-axis. 
Therefore $$\left\langle \sin^2(x) - \frac{1}{2} \right\rangle = 0$$    
Adding   $\displaystyle \frac{1}{2} $ to both sides of the equation yields the desired result.
A: 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,
$$\langle \sin^2 \omega t \rangle = \frac{1}{2}.$$
A: The valve of sine function is between 0 and 1 . So we know the average formula
Average = (Minimum+Maximum)÷2
So,
Average = (0+1)÷2
= 1/2
