## Thomas' Calculus 13th Edition

$-6 \sqrt 3$
Consider $f(x)=\int_{-(\pi/3)}^{(\pi/3)} (\sec^2 x-8 \cos x) dx$ Use formula such as: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ As we know that $\int_{-k}^k f(x) dx=2 \int_0^{k} f(x) dx$ This implies that $2[\tan x-8 \sin x]_{0}^{\pi/3}=2[(\tan (\dfrac{\pi}{3})-\tan (0))-8 (\sin (\dfrac{\pi}{3})-\sin (0))=2\sqrt 3-8\sqrt 3=\sqrt 3 (2-8)$ Hence, $f(x)=\int_{-(\pi/3)}^{(\pi/3)} (\sec^2 x-8 \cos x) dx=-6 \sqrt 3$