Average of cos(x)^2 in spherical distribution

[joris_pixie]

<< Mentor Note -- Thread moved from the technical forums so no Homework Help Template is shown >>

I'm particle physics at the moment and I don't get why the average value of cosˆ2 is 1/3.
The section :

My solution is :
$$< cos^{2}(\alpha)> = \frac{1}{\pi - 0} \int_{0}^{\pi } cos^{2}(\alpha)sin(\alpha)d\alpha$$
I substitute
$$u = cos(\alpha)$$
so i get
$$-\frac{1}{\pi}\int_{1}^{-1}u^{2}du = \frac{2}{3\pi}$$

What I am i doing wrong ? :(

 

[Ray Vickson]

<< Mentor Note -- Thread moved from the technical forums so no Homework Help Template is shown >>

I'm particle physics at the moment and I don't get why the average value of cosˆ2 is 1/3.
The section :
View attachment 204679

My solution is :
$$< cos^{2}(\alpha)> = \frac{1}{\pi - 0} \int_{0}^{\pi } cos^{2}(\alpha)sin(\alpha)d\alpha$$
I substitute
$$u = cos(\alpha)$$
so i get
$$-\frac{1}{\pi}\int_{1}^{-1}u^{2}du = \frac{2}{3\pi}$$

What I am i doing wrong ? :(

$$\langle \cos^2(\alpha) \rangle =
\frac{\int_0^{\pi} \cos^2(\alpha) \sin(\alpha) \, d \alpha}{\int_0^{\pi} 1 \cdot \sin(\alpha) \, d \alpha}. $$
To compute an average like $\langle f(x) \rangle = \int w(x) f(x) dx,$ you need the "weight" function $w(x) \geq 0$ to integrate to 1.

 

[joris_pixie]

$$\langle \cos^2(\alpha) \rangle =
\frac{\int_0^{\pi} \cos^2(\alpha) \sin(\alpha) \, d \alpha}{\int_0^{\pi} 1 \cdot \sin(\alpha) \, d \alpha}. $$
To compute an average like $\langle f(x) \rangle = \int w(x) f(x) dx,$ you need the "weight" function $w(x) \geq 0$ to integrate to 1.

Thank you ! Makes sense!