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

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

Vedclass · Accepted Answer

$2\pi$

Vedclass · Answer

(D) Let $I = \int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$. Expanding the integrand,we get: $I = \int_{-\pi}^{\pi} (\cos^2 px + \sin^2 qx - 2 \cos px \sin qx) dx$. Using the property $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,the term $\int_{-\pi}^{\pi} 2 \cos px \sin qx dx = 0$ because $\cos px$ is even and $\sin qx$ is odd. Thus,$I = \int_{-\pi}^{\pi} \cos^2 px dx + \int_{-\pi}^{\pi} \sin^2 qx dx$. Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ and $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$: $I = \int_{-\pi}^{\pi} \frac{1 + \cos 2px}{2} dx + \int_{-\pi}^{\pi} \frac{1 - \cos 2qx}{2} dx$. $I = \frac{1}{2} [x + \frac{\sin 2px}{2p}]_{-\pi}^{\pi} + \frac{1}{2} [x - \frac{\sin 2qx}{2q}]_{-\pi}^{\pi}$. Since $p, q$ are integers,$\sin(2p\pi) = 0$ and $\sin(-2p\pi) = 0$. $I = \frac{1}{2} [(\pi - 0) - (-\pi - 0)] + \frac{1}{2} [(\pi - 0) - (-\pi - 0)]$. $I = \frac{1}{2} [2\pi] + \frac{1}{2} [2\pi] = \pi + \pi = 2\pi$.

The value of the integral $\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$,where $p$ and $q$ are integers,is equal to:

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