How do you integrate $int x^2cos(3x)$ by integration by parts method?

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Answer 1 · 2017-02-03T10:13:23

$1/27(9x^2-2)sin3x+(2x)/9cos3x+C$

$I=intuv'dx=uv-intvu'dx$

$intx^2cos3xdx$

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

$v'=cos3x=>v=1/3sin3x$

$:.I_1=x^2/3sin3x-int[(2x)/3sin3x]dx$

$:.I_1=x^2/3sin3x-2/3color(blue)(int[xsin3x]dx)$

we now have to use IBP on the integral highlighted in blue

$I_2=(int[xsin3x]dx)$

$u=x=>u'=1$

$v'=sin3x=>v=-1/3cos3x$

$I_2=(-x/3cos3x-int-1/3cos3xdx)$

$I_2=-x/3cos3x+1/9sin3x$

substituting back into $I_1$

$:.I_1=x^2/3sin3x-2/3(-x/3cos3x+1/9sin3x)+C$

$:.I_1=x^2/3sin3x+(2x)/9cos3x-2/27sin3x+C$

$I_1=(x^2/3-2/27)sin3x+(2x)/9cos3x+C$

$I_1=1/27(9x^2-2)sin3x+(2x)/9cos3x+C$

How do you integrate int x^2cos(3x)int x^2cos(3x) by integration by parts method?