How do you integrate $int x^2*sin(2x) dx$ from $[0,pi/2]$ ?

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Answer 1 · 2018-04-12T16:13:22

The answer is $=pi^2/8-1/2$

First, calculate the indefinite integral by applying the integration by parts $2$ times

$intuv'=uv-intu'v$

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

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

The integral is

$I=intx^2sin2xdx=-x^2/2cos2x+intxcos2xdx$

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

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

$intxcos2xdx=x/2sin2x-1/2intsin2xdx$

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

Finally,

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

And the definite integral is

$intx^2sin2xdx=[-x^2/2cos2x+x/2sin2x+1/4cos2x]_0^(pi/2)$

$=(pi^2/8-1/4)-(1/4)$

$=pi^2/8-1/2$

$=0.73$

How do you integrate int x^2*sin(2x) dxint x^2*sin(2x) dx from [0,pi/2][0,pi/2]?