Question

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

Answer 1

\[Let I = \int_0^\frac{\pi}{2} x^2 \cos^2 x d x . Then, \]
\[I = \int_0^\frac{\pi}{2} x^2 \left( \frac{1 + \cos 2x}{2} \right)dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( \frac{x^2}{2} + \frac{x^2 \cos 2x}{2} \right) dx\]
\[ \Rightarrow I = \left[ \frac{x^3}{6} \right]_0^\frac{\pi}{2} + \left[ \frac{x^2 \sin 2x}{4} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} \frac{x}{2} \sin 2x\ d\ x\]
\[ \Rightarrow I = \left[ \frac{x^3}{6} \right]_0^\frac{\pi}{2} + \left[ \frac{x^2 \sin 2x}{4} \right]_0^\frac{\pi}{2} - \left[ \frac{- x \cos 2x}{4} \right]_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} - 1 \frac{\cos2x}{2}dx\]
\[ \Rightarrow I = \left[ \frac{x^3}{6} \right]_0^\frac{\pi}{2} + \left[ \frac{x^2 \sin 2x}{4} \right]_0^\frac{\pi}{2} + \left[ \frac{x \cos 2x}{4} \right]_0^\frac{\pi}{2} - \left[ \frac{\sin 2x}{4} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi^3}{48} - \frac{\pi}{8}\]