Answer
$\int x^{2} sin2x dx = \frac{2xsin2x + (1-2x^{2})cos2x}{4} +C $
Work Step by Step
The given indefinite integral is ,
$\int x^{2} sin2x dx$
Solving this with integration by parts below :
The formula of integration by parts is :
$\int udv=uv−\int vdu$
Now let's consider,
$u = x^{2}$
So, $du= 2xdx$
And ,
$dv= sin2x$
So, $v = -\frac{cos2x}{2}$
So, according to integration by parts,
$\int x^{2} sin2x dx = x^{2}(-\frac{cos2x}{2}) - \int (-\frac{cos2x}{2})2xdx$
⇒ $\int x^{2} sin2x dx = -\frac{ x^{2}cos2x}{2} + \int xcos2xdx .................(i)$
Now for $ \int xcos2xdx$ ,
Let's consider,
$u = x$
⇒ $du = dx$
And,
$dv= cos2x$
⇒ $v= \frac{sin2x}{2}$
So ,
$\int xcos2xdx = x \frac{sin2x}{2} - \int\frac{sin2x}{2}dx$
⇒ $\int xcos2xdx = \frac{xsin2x}{2} - \frac{1}{2}\int sin2xdx$
⇒ $\int xcos2xdx = \frac{xsin2x}{2} - \frac{1}{2}(-\frac{cos2x}{2})$
⇒ $\int xcos2xdx = \frac{xsin2x}{2} + \frac{cos2x}{4}$
Now put the value of $\int xcos2xdx$ in equation $(i)$ ,
$\int x^{2} sin2x dx = -\frac{ x^{2}cos2x}{2} + \frac{xsin2x}{2} + \frac{cos2x}{4} +C $
⇒ $\int x^{2} sin2x dx = \frac{xsin2x}{2} + \frac{cos2x}{4} -\frac{ x^{2}cos2x}{2} +C $
⇒ $\int x^{2} sin2x dx = \frac{2xsin2x + cos2x-2x^{2}cos2x}{4} +C $
⇒ $\int x^{2} sin2x dx = \frac{2xsin2x + (1-2x^{2})cos2x}{4} +C $
