int_{-pi}^{pi} (cos px - sin qx)^2 dx is equa | vedclass

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(D) Let $I = \int_{-\pi}^{\pi} (\cos^2 px + \sin^2 qx - 2 \sin qx \cos px) dx$. Using the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if

Vedclass · Accepted Answer

$2\pi$

Vedclass · Answer

(D) Let $I = \int_{-\pi}^{\pi} (\cos^2 px + \sin^2 qx - 2 \sin qx \cos px) dx$. Using the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(x)$ is even and $0$ if $f(x)$ is odd: $I = \int_{-\pi}^{\pi} \cos^2 px dx + \int_{-\pi}^{\pi} \sin^2 qx dx - 2 \int_{-\pi}^{\pi} \sin qx \cos px dx$. Since $\sin qx \cos px$ is an odd function,its integral over $[-\pi, \pi]$ is $0$. $I = \int_{-\pi}^{\pi} \frac{1 + \cos 2px}{2} dx + \int_{-\pi}^{\pi} \frac{1 - \cos 2qx}{2} dx$. $I = \frac{1}{2} [x + \frac{\sin 2px}{2p}]_{-\pi}^{\pi} + \frac{1}{2} [x - \frac{\sin 2qx}{2q}]_{-\pi}^{\pi}$. $I = \frac{1}{2} [(\pi - (-\pi)) + 0] + \frac{1}{2} [(\pi - (-\pi)) - 0] = \frac{1}{2} (2\pi) + \frac{1}{2} (2\pi) = \pi + \pi = 2\pi$.

$\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$ is equal to (where $p$ and $q$ are integers).

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