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

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

Vedclass · Accepted Answer

$\frac{1}{{36}}({\pi ^2} + 16)$

Vedclass · Answer

(D) Let $I = \int_0^{\pi /6} {(2 + 3{x^2})\cos 3x\,dx}$.Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = 2 + 3x^2$ 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^{\pi /6} - \int_0^{\pi /6} \frac{\sin 3x}{3} (6x) \, dx$$I = \left[ (2 + 3(\frac{\pi^2}{36})) \frac{\sin(\pi/2)}{3} - 0 \right] - 2 \int_0^{\pi /6} x \sin 3x \, dx$$I = \frac{1}{3} (2 + \frac{\pi^2}{12}) - 2 \left[ x \left( -\frac{\cos 3x}{3} \right) - \int_0^{\pi /6} (1) \left( -\frac{\cos 3x}{3} \right) \, dx \right]$$I = \frac{2}{3} + \frac{\pi^2}{36} - 2 \left[ -\frac{x \cos 3x}{3} + \frac{1}{3} \int_0^{\pi /6} \cos 3x \, dx \right]$$I = \frac{2}{3} + \frac{\pi^2}{36} + \frac{2x \cos 3x}{3} - \frac{2}{3} \left[ \frac{\sin 3x}{3} \right]_0^{\pi /6}$Evaluating at limits: $I = \frac{2}{3} + \frac{\pi^2}{36} + \frac{2}{3} (\frac{\pi}{6} \cos(\pi/2)) - \frac{2}{9} (\sin(\pi/2) - \sin 0)$Since $\cos(\pi/2) = 0$ and $\sin(\pi/2) = 1$,$I = \frac{2}{3} + \frac{\pi^2}{36} + 0 - \frac{2}{9} = \frac{6-2}{9} + \frac{\pi^2}{36} = \frac{4}{9} + \frac{\pi^2}{36} = \frac{16 + \pi^2}{36} = \frac{1}{36}(\pi^2 + 16)$.

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

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