The value of the integral int_{-pi}^{pi} (cos | vedclass

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(D) Let $I = \int_{-\pi}^{\pi} (\cos ax - \sin bx)^2 dx$.Expanding the integrand,we get $I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx - 2\cos ax \sin

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$2\pi$

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(D) Let $I = \int_{-\pi}^{\pi} (\cos ax - \sin bx)^2 dx$.Expanding the integrand,we get $I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx - 2\cos ax \sin bx) dx$.Using the linearity of the integral,$I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx) dx - 2\int_{-\pi}^{\pi} \cos ax \sin bx dx$.Since $\cos ax \sin bx$ is an odd function,the integral $\int_{-\pi}^{\pi} \cos ax \sin bx dx = 0$.Thus,$I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx) dx$.Since the integrand is an even function,$I = 2 \int_{0}^{\pi} (\cos^2 ax + \sin^2 bx) dx$.Using the identities $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ and $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we have:$I = 2 \int_{0}^{\pi} (\frac{1 + \cos 2ax}{2} + \frac{1 - \cos 2bx}{2}) dx = \int_{0}^{\pi} (2 + \cos 2ax - \cos 2bx) dx$.Integrating term by term,$I = [2x + \frac{\sin 2ax}{2a} - \frac{\sin 2bx}{2b}]_{0}^{\pi}$.Since $a$ and $b$ are integers,$\sin(2a\pi) = 0$ and $\sin(2b\pi) = 0$.Therefore,$I = 2\pi + 0 - 0 = 2\pi$.

The value of the integral $\int_{-\pi}^{\pi} (\cos ax - \sin bx)^2 dx$,where $a$ and $b$ are integers,is

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