Evaluation of

$$\int_0^{\pi/2} \cos^4 \psi \, d \psi~:$$
$$\int_0^{\pi/2} \cos^4 \psi \, d \psi = \int_0^{\pi/2} \cos^2 \psi (1 - \sin^2 \psi ) \, d \psi$$
$$\int_0^{\pi/2} \cos^4 \psi \, d \psi = \int_0^{\pi/2} \cos^2 \psi \, d \psi - \int_0^{\pi/2} \cos^2 \psi \sin^2 \psi \, d \psi$$
We already found that $\langle \cos^2 \psi \rangle = 1/2$ so $\int_0^{\pi/2} \cos^2 \psi \, d \psi = \pi / 4$. Integrate the remaining integral by parts using $$u \equiv \cos^2 \psi \sin \psi {\rm ~and~} dv \equiv \sin \psi \, d \psi$$
therefore $$du = \cos^3 \psi - 2\sin^2\psi\cos\psi {\rm ~and~} v = -\cos\psi$$ and $$\int_0^{\pi/2} \cos^2 \psi \sin^2 \psi \, d \psi = -\cos^2 \psi \sin \psi \cos \psi \vert_0^{\pi/2} - $$ $$                   \int_0^{\pi/2} -\cos \psi ( \cos^3 \psi - 2 \sin^2 \psi \cos \psi ) d \psi$$
$$\int_0^{\pi/2} \cos^2 \psi \sin^2 \psi \, d \psi = \int_0^{\pi/2} \cos^4 \psi \, d \psi - 2\int_0^{\pi/2} \cos^2 \psi \sin^2 \psi \, d \psi$$ which has the same integral on both sides, so
$$\int_0^{\pi/2} \cos^2 \psi \sin^2 \psi \, d \psi = {1 \over 3} \int_0^{\pi/2} \cos^4 \psi \, d \psi$$
Using the relation from the 3rd line of this evaluation we get

$$\int_0^{\pi/2} \cos^4 \psi \, d \psi = {\pi \over 4} - {1 \over 3} \int_0^{\pi/2} \cos^4 \psi \, d \psi$$
$${4 \over 3} \int_0^{\pi/2} \cos^4 \psi \, d \psi = {\pi \over 4}$$
so
$$\int_0^{\pi/2} \cos^4 \psi \, d \psi = {3 \pi \over 16}$$