Integral Table

\[\int A x^2 \sin^2(kx) \, dx = \frac{A}{6} x^3 - \frac{A}{4k^2} x^2 \sin(2kx) - \frac{A}{4k^3} x \cos(2kx) + \frac{A}{8k^3} \sin(2kx) + C.\]

Definite Integrals

\[\int_0^\pi \cos^2(x) \, dx = \frac{\pi}{2}\]

The area under \(\cos\) and \(\cos^2\) is \(\pi\) times the amplitude. Since our amplitude is only \(1\) and the range is only have of a full cycle, the area under is \(\pi/2\). So for a full cycle, it should be twice a half-cycle, or \(\pi\), so,

\[\int_0^{2\pi} \cos^2(x) \, dx = \pi.\]