I'm trying to figure out how to calculate the energy content of a signal over a single period. This is a problem from a workbook.
The signal in question is the following $x(t) = t\cdot \cos(\omega\cdot t)$
I understand that I need to calculate the following integral.
$\int_{-T_0/2}^{T_0/2}|t\cdot \cos(\omega t)|^2 dt$ where $T_0 = \frac{2\pi}{\omega}$
After a lot of calculus and help from wolfram alpha I learn the indefinite integral is
$$\int_{-T_0/2}^{T_0/2}|t\cdot \cos(\omega t)|^2 dt = $$
$$\frac{4 t^3 \omega^3 + (6 t^2 \omega^2 - 3)\cdot \sin(2 t \omega) + 6 t \omega \cdot \cos(2 t \omega)}{(24 \omega^3)} $$
I next evaluate this integral from $-\frac{T_0}{2}$ to $\frac{T_0}{2}$ and I get
$$\frac{1}{\omega^3}\cdot \frac{2\pi^3 + 3\pi}{6}$$
The book's answer is $$\frac{\pi}{2\cdot\omega^3}$$
I don't know wheter I went wrong. It may be in terms of calculus, algebra, or my initial assumptions as to calculating energy over a single period. Any help would be greatly appreciated.
If you wanted to find the integral of square (it agrees with the definition on wiki), then your result is correct. You should check, however, if you need to divide it by the length of the period.
Make a change of variables $u = t\omega$, then use integration by parts: $$\int_{-\pi/\omega}^{\pi/\omega} t^2\cos^2(\omega t) dt = \frac{1}{\omega^3}\int_{-\pi}^{\pi} u^2\cos^2(u) du$$ $$ = \frac{1}{2\omega^3}\int_{-\pi}^{\pi} u^2(\cos 2u +1) du$$ $$=\frac{\pi^3}{3\omega^3} + \frac{1}{2\omega^3}\int_{-\pi}^{\pi} u^2 \cos 2u\, du$$
$$= \frac{\pi^3}{3\omega^3} + \frac{1}{2\omega^3}\left(\frac 12u^2\sin 2u \big|_{-\pi}^{\pi} - \int_{-\pi}^{\pi}u \sin 2u \,du\right)$$
$$= \frac{\pi^3}{3\omega^3} - \frac{1}{2\omega^3}\left( \int_{-\pi}^{\pi}u \sin 2u \,du\right)$$ $$= \frac{\pi^3}{3\omega^3} - \frac{1}{2\omega^3}\left( -\frac 12 u \cos 2u\big|_{-\pi}^{\pi} + \frac 12\int_{-\pi}^{\pi}\cos 2u \,du\right)$$ $$= \frac{\pi^3}{3\omega^3} + \frac{ \pi}{2\omega^3} .$$