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(D) We know that $\sin^2 x = \frac{1 - \cos 2x}{2}$.Substituting this into the integral:$\int x \sin^2 x \, dx = \int x \left( \frac{1 - \cos 2x}{2} \

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$\frac{x^2}{4} - \frac{x}{4} \sin 2x - \frac{1}{8} \cos 2x + c$

Vedclass · Answer

(D) We know that $\sin^2 x = \frac{1 - \cos 2x}{2}$.Substituting this into the integral:$\int x \sin^2 x \, dx = \int x \left( \frac{1 - \cos 2x}{2} \right) dx$$= \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos 2x \, dx$$= \frac{1}{2} \left( \frac{x^2}{2} \right) - \frac{1}{2} \left[ x \left( \frac{\sin 2x}{2} \right) - \int 1 \cdot \frac{\sin 2x}{2} \, dx \right]$$= \frac{x^2}{4} - \frac{x}{4} \sin 2x + \frac{1}{2} \int \frac{\sin 2x}{2} \, dx$$= \frac{x^2}{4} - \frac{x}{4} \sin 2x + \frac{1}{4} \left( -\frac{\cos 2x}{2} \right) + c$$= \frac{x^2}{4} - \frac{x}{4} \sin 2x - \frac{1}{8} \cos 2x + c$.

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

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