Question

\[\int\limits_0^{\pi/4} x^2 \sin\ x\ dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{4} x^2 \sin\ x\ d x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ - x^2 \cos x \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} - 2x \cos\ x\ d\ x\]
\[ \Rightarrow I = \left[ - x^2 \cos x \right]_0^\frac{\pi}{4} + \left[ 2x \sin x \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} 2 \sin\ x\ dx\]
\[ \Rightarrow I = \left[ - x^2 \cos x \right]_0^\frac{\pi}{4} + \left[ 2x \sin x \right]_0^\frac{\pi}{4} + \left[ 2 \cos x \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow I = \frac{- \pi^2}{16\sqrt{2}} + \frac{\pi}{2\sqrt{2}} + \frac{2}{\sqrt{2}} - 2\]
\[ \Rightarrow I = \sqrt{2} + \frac{\pi}{2\sqrt{2}} - \frac{\pi^2}{16\sqrt{2}} - 2\]