$\int\limits_{-\pi}^\pi(\cos a x-\sin b x)^2$ (where a, b are integers) =

$-\pi$ 0 $\pi$ $2 \pi$

$2 \pi$

$I=\int\limits_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x-2 \int\limits_{-\pi}^\pi \cos a x \sin a x d x$ $=2 \int\limits_0^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x-0$ $=\int\limits_0^\pi(1+\cos 2 a x+1-\cos 2 b x) d x=2 \pi+0=2 \pi$ Hence (4) is the correct answer.