$\int x^{2} \sin x \cos x d x =$

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$\begin{aligned} \frac{x^{2} \cos 2 x}{2} + \frac{x \sin 2 x}{4} + \frac{\cos 2 x}{8} + c \end{aligned}$

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

$\begin{aligned} \frac{(1 - 2 x)^{2}}{2} \cos 2 x + x \sin 2 x + c \end{aligned}$

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

answer is C.

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Detailed Solution

$\begin{array}{l} \int x^{2} \sin x \cos x d x = \frac{1}{2} \int x^{2} (2 \sin x \cos x) d x = \frac{1}{2} \int x^{2} \cdot \sin 2 x d x \\ = \frac{1}{2} [x^{2} (- \frac{1}{2} \cos 2 x) - \int (- \frac{1}{2} \cos 2 x) (2 x) d x] = \frac{- 1}{4} x^{2} \cos 2 x + \frac{1}{2} \int x \cos 2 x d x \\ = \frac{- 1}{4} x^{2} \cos 2 x + \frac{1}{2} [x \frac{\sin 2 x}{2} - \int \frac{\sin 2 x}{2} d x] = \frac{- 1}{4} x^{2} \cos 2 x + \frac{1}{4} x \sin 2 x + \frac{1}{8} \cos 2 x + c \\ = \frac{(1 - 2 x^{2})}{8} \cos 2 x + \frac{x}{4} \sin 2 x + c . \end{array}$