Why does this approximation of integral of cosine work?

Question

I read this out of my textbook that if $f_cT \gg 1$, then $${1\over T}\int_0^T(\cos2\pi f_c\tau ) + \cos(4\pi f_ct-2\pi f_c\tau)dt \approx \cos(2\pi f_c\tau ).$$ If this is true, then I think it's equivalent to say $${1\over T}\int_0^T\cos(4\pi f_ct-2\pi f_c\tau)dt \approx 0.$$ However I don't know why this is true, so could someone help me to understand this?

Answer 1

Think of the second integral as $\int_{0}^{T} \cos(at+b)dt$, and just evaluate it.

Modulo a typo, we get:

$|\frac{1}{T} \int_0^T\cos(4\pi f_ct-2 \pi f_c \tau)dt| = |\frac{\sin(4\pi f_c T-2 \pi f_c \tau)-\sin(-2 \pi f_c \tau)}{4\pi f_cT}| \le |\frac{2}{4\pi f_cT}|$,

and you have your result.