Evaluate: ^ _ - ( ax - bx)^ 2 dx - Vidyadip

Question

$\text{Evaluate}: \int\limits^{\pi}_{-\pi} (\cos ax - \sin bx)^{2} dx$

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Answer

$\text{I} = \int\limits^{\pi}_{-\pi} (\cos \text{ax} - \sin \text{bx})^{2} \text{dx}=\int\limits^{\pi}_{-\pi} (\cos^{2} \text{ax} - \sin^{2} \text{bx}) \text{dx} - \int\limits^{\pi}_{-\pi} 2\cos \text{ax} \sin \text{bx dx}$
$\text{I}_{1} = 2 \int\limits^{\pi}_{0}(\cos^{2}\text{ax} + \sin^{2}\text{bx})\text{dx} \text{(being an even fun.)}$
$\text{I}_{2} = \text{0 (being an odd fun.)}$
$\therefore \text{I = I}_{1} = \int\limits^{\pi}_{0}( 1 + \cos \text{2ax} + 1 -\cos\text{2bx}) \text{dx}$
$= \bigg[2\text{x} + \frac{\sin \text{2ax}}{\text{2a}} - \frac{\sin \text{2bx}}{\text{2b}} \bigg]^{\pi}_{0}$
$= \bigg[2{\pi} + \frac{1}{\text{2a}} . \sin \text{2a}\pi - \frac{\sin 2\text{b}{\pi}}{2\text{b}}\bigg]\text{or 2} \pi$

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