Question

\[\int x^2 \sin^2 x\ dx\]

Answer 1

\[\int x^2 \sin^2 x\ dx\]
`   " Taking x"^2"  as the first function and sin"^2 x " as the second function . " ` 
\[ = x^2 \int\frac{1 - \cos2x}{2} - \int\left\{ \frac{d}{dx}\left( x^2 \right)\int\frac{1 - \cos2x}{2}dx \right\}dx\]

` = x^2/2  ( x - {sin 2x}/2 ) - ∫   x^2dx + ∫  { x sin 2x} /2 dx `

  `[   \text{  Here, taking x as the first function and sin 2x as the second function} ]. ` 
`=  x^3 / 2 - { x^2 sin 2x}/4   - x^3/3 + 1/2 [ x  ∫  sin 2x - ∫  { d /dx (x) ∫   sin  2x  dx } dx] `

\[ = \frac{x^3}{2} - \frac{x^2 \sin2x}{4} - \frac{x^3}{3} + \frac{1}{2}\left[ \frac{- x\cos2x}{2} + \int\frac{\text{ cos 2x  dx }}{4} \right]\]
\[ = \frac{x^3}{6} - \frac{x^2 \sin2x}{4} - \frac{x \cos2x}{4} + \frac{\sin2x}{8} + C\]