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(B) To evaluate $I = \int x^2 \sin 2x \, dx$,we use the integration by parts formula: $\int u \, dv = uv - \int v \, du$. Let $u = x^2$ and $dv = \sin

Vedclass · Accepted Answer

$-\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x \sin 2x + \frac{1}{4} \cos 2x + c$

Vedclass · Answer

(B) To evaluate $I = \int x^2 \sin 2x \, dx$,we use the integration by parts formula: $\int u \, dv = uv - \int v \, du$. Let $u = x^2$ and $dv = \sin 2x \, dx$. Then $du = 2x \, dx$ and $v = -\frac{\cos 2x}{2}$. Applying the formula: $I = x^2 \left(-\frac{\cos 2x}{2}\right) - \int \left(-\frac{\cos 2x}{2}\right) (2x \, dx)$ $I = -\frac{x^2 \cos 2x}{2} + \int x \cos 2x \, dx$. Now,apply integration by parts again for $\int x \cos 2x \, dx$: Let $u = x$ and $dv = \cos 2x \, dx$. Then $du = dx$ and $v = \frac{\sin 2x}{2}$. $\int x \cos 2x \, dx = x \left(\frac{\sin 2x}{2}\right) - \int \frac{\sin 2x}{2} \, dx$ $= \frac{x \sin 2x}{2} - \frac{1}{2} \left(-\frac{\cos 2x}{2}\right) = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4}$. Substituting this back into the expression for $I$: $I = -\frac{x^2 \cos 2x}{2} + \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} + c$.

$\int x^2 \sin 2x \, dx = $

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