int_0^{frac{pi}{6}} (2+3x^2) cos 3x , dx = | vedclass

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(B) Let $I = \int_0^{\frac{\pi}{6}} (2+3x^2) \cos 3x \, dx$. Using integration by parts $\int u \, dv = uv - \int v \, du$,let $u = 2+3x^2$ and $dv =

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$\frac{4}{9} + \frac{\pi^2}{36}$

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(B) Let $I = \int_0^{\frac{\pi}{6}} (2+3x^2) \cos 3x \, dx$. Using integration by parts $\int u \, dv = uv - \int v \, du$,let $u = 2+3x^2$ and $dv = \cos 3x \, dx$. Then $du = 6x \, dx$ and $v = \frac{\sin 3x}{3}$. $I = \left[ (2+3x^2) \frac{\sin 3x}{3} \right]_0^{\frac{\pi}{6}} - \int_0^{\frac{\pi}{6}} \frac{\sin 3x}{3} (6x) \, dx$. $I = \left[ (2+3(\frac{\pi^2}{36})) \frac{\sin(\frac{\pi}{2})}{3} - 0 \right] - 2 \int_0^{\frac{\pi}{6}} x \sin 3x \, dx$. $I = \frac{1}{3} (2 + \frac{\pi^2}{12}) - 2 \int_0^{\frac{\pi}{6}} x \sin 3x \, dx$. For the second integral,use integration by parts again: $u = x, dv = \sin 3x \, dx \implies du = dx, v = -\frac{\cos 3x}{3}$. $\int_0^{\frac{\pi}{6}} x \sin 3x \, dx = \left[ -\frac{x \cos 3x}{3} \right]_0^{\frac{\pi}{6}} - \int_0^{\frac{\pi}{6}} -\frac{\cos 3x}{3} \, dx$. $= (0 - 0) + \frac{1}{3} \left[ \frac{\sin 3x}{3} \right]_0^{\frac{\pi}{6}} = \frac{1}{9} \sin(\frac{\pi}{2}) = \frac{1}{9}$. Substituting back: $I = \frac{2}{3} + \frac{\pi^2}{36} - 2(\frac{1}{9}) = \frac{2}{3} - \frac{2}{9} + \frac{\pi^2}{36} = \frac{6-2}{9} + \frac{\pi^2}{36} = \frac{4}{9} + \frac{\pi^2}{36}$.

$\int_0^{\frac{\pi}{6}} (2+3x^2) \cos 3x \, dx =$

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