Question

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

Answer 1

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