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 $$u \equiv \cos^2 \psi \sin \psi {\rm ~and~} dv
\equiv \sin \psi d \psi$$
$$\int_0^{\pi/2} \cos^2 \psi \sin^2
\psi d \psi = -\cos^2 \psi \sin
\psi \cos \psi \vert_0^{\pi/2} - $$
$$
\int_0^{\pi/2} -\cos \psi ( \cos^3
\psi - \sin^2 \psi \times 2 \cos \psi ) d \psi$$
$$\int_0^{\pi/2} \cos^2 \psi \sin^2
\psi d \psi = \int_0^{\pi/2}
\cos^4 \psi d \psi - 2\int_0^{\pi/2} \cos^2 \psi \sin^2 \psi d \psi$$
$$\int_0^{\pi/2} \cos^2 \psi \sin^2
\psi d \psi = {1 \over 3}
\int_0^{\pi/2} \cos^4 \psi d \psi$$
$$\int_0^{\pi/2} \cos^4 \psi d \psi
= {\pi \over 4} - {1 \over 3}
\int_0^{\pi/2} \cos^4 \psi d \psi$$
$${4 \over 3} \int_0^{\pi/2} \cos^4
\psi d \psi = {\pi \over 4}$$
so
$$\int_0^{\pi/2} \cos^4 \psi d \psi
= {3 \pi \over 16}$$