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Question
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
Solution
\[\int_0^\frac{\pi}{2} x^2 \cos2x d x\]
\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2x \frac{\sin2x}{2}dx\]
\[ = \left[ x^2 \frac{sin2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} x \sin 2x dx\]
\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} - \left[ - x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} + \left[ - \int_0^\frac{\pi}{2} \frac{\cos2x}{2}dx \right]\]
\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} + \left[ x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} \frac{\cos2x}{2}dx\]
\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} + \left[ x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} - \frac{1}{2} \left[ \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]
\[ = 0 - \frac{\pi}{4} - 0\]
\[ = \frac{- \pi}{4}\]
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