Question

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

Answer 1

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