Trigonometric integrals by linearisation

First.
$$\int_0^{\pi/2}\cos^2t\,dt=\frac12\left[t+\frac{\sin2t}2\right]_0^{\pi/2}=\frac\pi4.$$

Second.
$$\int_0^{\pi/3}(1-\cos3x)\,dx=\left[x-\frac{\sin3x}3\right]_0^{\pi/3}=\frac\pi3-0=\frac\pi3.$$

Third. $\sin^4x=\left(\frac{1-\cos2x}2\right)^2=\frac14\left(1-2\cos2x+\cos^22x\right)$ and $\cos^22x=\frac{1+\cos4x}2$, so
$$\sin^4x=\frac38-\frac12\cos2x+\frac18\cos4x.$$
The two cosine terms integrate to $0$ over $[0,\pi]$, hence $\int_0^\pi\sin^4x\,dx=\frac{3\pi}8$.

Remark. The average value of $\sin^2$ is $\frac12$ and that of $\sin^4$ is $\frac38$: these are the first Wallis numbers.