Question

\[\int\limits_0^{\pi/2} x^2 \cos\ 2x\ dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{2} x^2 \cos 2x d x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ x^2 \frac{\sin 2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2x \frac{\sin 2x}{2} d x\]
\[ \Rightarrow I = \left[ x^2 \frac{\sin 2x}{2} \right]_0^\frac{\pi}{2} - \left[ - x \frac{\cos 2x}{2} \right]_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} - 1 \frac{\cos 2x}{2} d x\]
\[ \Rightarrow I = \left[ x^2 \frac{\sin 2x}{2} \right]_0^\frac{\pi}{2} + \left[ x \frac{\cos 2x}{2} \right]_0^\frac{\pi}{2} - \left[ \frac{\sin 2x}{4} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = 0 - \frac{\pi}{4} - 0\]
\[ \Rightarrow I = - \frac{\pi}{4}\]