a) \(\int \sin^{2}{x}dx\)
$$\int \sin^{2}{x} dx = \int\frac{1}{2}[1-\cos(2x)]dx = \frac{x}{2} - \frac{sin(2x)}{4} + C$$
Maneira II:
$$\int \sin(mx)\sin(nx) dx = \int \frac{1}{2}\left[\cos((m-n)x) - \cos((m+n)x)\right] dx, n = m = 1$$
$$\int \sin^2(x) dx = \int \frac{1}{2}\left[\underbrace{\cos(0)}_{1} - \cos(2x)\right] dx$$
$$ = \frac{1}{2}x + C - \frac{1}{2}\int\cos(2x)dx$$
substituindo u = 2x:
$$\frac{1}{2}x + C - \frac{1}{2}\underbrace{\int\frac{\cos(u)}{2}du}_{sin(u)/2}$$
Resolvendo a integral e trocando u = 2x:
$$\frac{1}{2}x - \frac{sin(2x)}{4} + C $$