I=∫x2sin2x dx=∫x21−cos2x2.dx=∫x22dx−∫x22cos2x dx=x36−x2sin2x4+12∫ddxx2sin2x2dx=x36−x2sin2x4+12∫2xsin2x2dx=x36−x24sin2x+12x(−cos2x)2−∫1(−cos2x2)dx=x36−x24sin2x−14xcos2x+sin2x4+c=x36−12[x22sin2x+xcos2x2−sin2x4]+c

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

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$\frac{1}{6} x^{3} - \frac{1}{2} [\frac{x^{2} \sin 2 x}{2} + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4}] + c$

$\frac{1}{6} x^{3} + \frac{1}{2} [\frac{x^{2} \sin 2 x}{2} + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4}] + c$

$\frac{1}{6} x^{3} - \frac{1}{2} [\frac{x^{2} \sin 2 x}{2} - \frac{x \cos 2 x}{2} + \frac{\sin 2 x}{4}] + c$

$- \frac{1}{6} x^{3} + \frac{1}{2} [\frac{x^{2} \sin 2 x}{2} + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4}] + c$

answer is D.

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Detailed Solution

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

$= \int x^{2} (\frac{1 - \cos 2 x}{2}) . d x$

$= \int \frac{x^{2}}{2} d x - \int \frac{x^{2}}{2} \cos 2 x d x$

$= \frac{x^{3}}{6} - \frac{x^{2} \sin 2 x}{4} + \frac{1}{2} \int \frac{d}{d x} x^{2} \frac{\sin 2 x}{2} d x$

$\begin{array}{l} = \frac{x^{3}}{6} - x^{2} \frac{\sin 2 x}{4} + \frac{1}{2} \int 2 x \frac{\sin 2 x}{2} d x \\ = \frac{x^{3}}{6} - \frac{x^{2}}{4} \sin 2 x + \frac{1}{2} (\frac{x (- \cos 2 x)}{2} - \int 1 (\frac{- \cos 2 x}{2}) d x) \end{array}$