How do you integrate $int x^2 sin^2 x dx$ using integration by parts?

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Answer 1 · 2018-03-20T15:57:19

The answer is $=1/6x^3-x^2/4sin2x-x/4cos2x-1/8sin2x+C$

Reminder

$cos2x=1-2sin^2x$

$sin^2x=1/2(1-cos2x)$

$intuv'=uv-intu'v$

The integral is

$I=intx^2sin^2xdx=1/2intx^2(1-cos2x)dx$

$=1/2intx^2-1/2intx^2cos2xdx$

$=1/6x^3-1/2intx^2cos2xdx$

Perform the second integral by parts

$u=x^2$ , $=>$ , $u'=2x$

$v'=cos2x$ , $=>$ , $v=1/2sin2x$

So,

$intx^2cos2xdx=x^2/2sin2x-intxsin2xdx$

Calculate the third intehral by parts

$u=x$ , $=>$ , $u'=1$

$v'=sin2x$ , $=>$ , $v=-1/2cos2x$

So,

$intxsin2xdx=-x/2cos2x+1/2intcos2xdx$

$=-x/2cos2x+1/4sin2x$

Putting it all together

$I=1/6x^3-x^2/4sin2x-x/4cos2x-1/8sin2x+C$

How do you integrate int x^2 sin^2 x dx int x^2 sin^2 x dx using integration by parts?