int_0^{pi / 4} x^2 sin 2x , dx = | vedclass

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(C) To evaluate the integral $I = \int_0^{\pi / 4} x^2 \sin 2x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = x^2$ an

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$\frac{\pi-2}{8}$

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(C) To evaluate the integral $I = \int_0^{\pi / 4} x^2 \sin 2x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = x^2$ and $dv = \sin 2x \, dx$. Then $du = 2x \, dx$ and $v = -\frac{\cos 2x}{2}$.$I = \left[ -\frac{x^2 \cos 2x}{2} \right]_0^{\pi / 4} - \int_0^{\pi / 4} \left( -\frac{\cos 2x}{2} \right) (2x) \, dx$$I = \left[ -\frac{x^2 \cos 2x}{2} \right]_0^{\pi / 4} + \int_0^{\pi / 4} x \cos 2x \, dx$Evaluating the first term: $-\frac{(\pi/4)^2 \cos(\pi/2)}{2} - 0 = 0$ (since $\cos(\pi/2) = 0$).Now evaluate $\int_0^{\pi / 4} x \cos 2x \, dx$ using integration by parts again:Let $u = x$ and $dv = \cos 2x \, dx$. Then $du = dx$ and $v = \frac{\sin 2x}{2}$.$\int_0^{\pi / 4} x \cos 2x \, dx = \left[ \frac{x \sin 2x}{2} \right]_0^{\pi / 4} - \int_0^{\pi / 4} \frac{\sin 2x}{2} \, dx$$= \left( \frac{(\pi/4) \sin(\pi/2)}{2} - 0 \right) - \left[ -\frac{\cos 2x}{4} \right]_0^{\pi / 4}$$= \frac{\pi}{8} + \left[ \frac{\cos 2x}{4} \right]_0^{\pi / 4} = \frac{\pi}{8} + \left( \frac{\cos(\pi/2)}{4} - \frac{\cos 0}{4} \right)$$= \frac{\pi}{8} + (0 - \frac{1}{4}) = \frac{\pi-2}{8}$.

$\int_0^{\pi / 4} x^2 \sin 2x \, dx =$

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