Simple integral leads to Kronecker delta term?

[the_dialogue]

Homework Statement

$$\int_{0}^{b} \int_{0}^{2\pi} C_{k,m}(r)^2 \left{\begin{array}{cc}cos(m\theta)^2\\sin(m\theta)^2 \end{array}\right} r dr d\theta$$

Homework Equations

See above

The Attempt at a Solution

Ignoring the 'r' integral for a second, the solution that I see written here is:
$$\pi(1+\delta_{0,m}(\int_{0}^{b} C_{k,m}(r)^2 r dr)$$
where $$\delta$$ is Kronecker delta.

Where did the tex]\pi(1+\delta_(0,m)}[/tex] come from?

I suppose my confusion may strand from misunderstanding what the bracketed term $$\left{\begin{array}{cc}cos(m\theta)^2\\sin(m\theta)^2 \end{array}\right}$$ means. What is this notation? Instinct says column vector, but that of course doesn't seem to be the case.

Thank you for any help!

 

[fzero]

The correct formula is more like

$$\pi(1+\delta_{0,m} ) (\int_{0}^{b} C_{k,m}(r)^2 r dr)$$

The factor in front is the result of doing the trig integration:

$$\int_0^{2\pi} \cos^2(n\phi) d\phi = \int_0^{2\pi} \sin^2(n\phi) d\phi =\pi, ~\text{for}~n=1,2,\ldots,$$
$$\int_0^{2\pi} \cos^2(0\phi) d\phi = 2\pi.$$

 

[the_dialogue]

The correct formula is more like

$$\pi(1+\delta_{0,m} ) (\int_{0}^{b} C_{k,m}(r)^2 r dr)$$

The factor in front is the result of doing the trig integration:

$$\int_0^{2\pi} \cos^2(n\phi) d\phi = \int_0^{2\pi} \sin^2(n\phi) d\phi =\pi, ~\text{for}~n=1,2,\ldots,$$
$$\int_0^{2\pi} \cos^2(0\phi) d\phi = 2\pi.$$

Of course!

My mistake. Thanks fzero for the prompt response.