OP Here -
Thanks to Semiclassical's answer I've decided to answer my own questions.
(Also for more insight on the trig identities check out this)
So we have $$\int_{t_0}^{t_0 +T}\cos(\frac{2\pi}{T}kt)\cos(\frac{2\pi}{T}mt)$$
We will use the following trigonometric formula $$\cos(A)\cos(B) = \frac{1}{2}\left(cos(A+B) + cos(A - B)\right)$$
Applying that identity
$$\int_{t_0}^{t_0 +T}\cos(\frac{2\pi}{T}kt)\cos(\frac{2\pi}{T}mt)dt = \frac{1}{2}\int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k+m)) + \cos(\frac{2\pi t}{T}(k-m))\right)dt$$
$$ =\frac{1}{2} \left( \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k+m))\right)dt + \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k-m))\right)dt \right)$$
Case 1: $k = m$
$$ = \frac{1}{2} \left( \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(2k))\right)dt + \int_{t_0}^{t_0 +T}\cos(0)dt \right)$$
Notice that the first term contributes $0$ because we are integrating cosine over an entire period and the "positive areas" and "negative areas" negate each other, while the second term is integrating $1$ over the period. Thus,
$$ = \frac{1}{2} \left( \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(2k))\right)dt + \int_{t_0}^{t_0 +T}\cos(0)dt \right) \\ = \frac{1}{2} \int_{t_0}^{t_0 +T}dt = \frac{T}{2}$$
Case 2: $k \neq m$
From
$$=\frac{1}{2} \left( \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k+m))\right)dt + \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k-m))\right)dt \right)$$
Both terms contribute $0$ since they are integrating a cosine wave over a full period with "negative" and "positive" areas
And of course, a similar argument can be made for sine waves.
