How do you integrate $int x^2 cos^2 3 x dx$ using integration by parts?

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Answer 1 · 2016-02-29T07:35:52

$x^2(x/2-sin(6x)/12) -x^3/3 - x cos(6x) + sin(6x)/6$

Please check calculation too

$u = x^2$
$cos^2 3x dx= {1 + cos(6x)}/2 dx= dv$

$du = 2x dx$
$v = x/2 - sin(6x)/12$

integral $I = x^2(x/2-sin(6x)/12) -x^3/3 + int (x/6 sin(6x))dx$

integrate by parts again
$u=x$
$du=dx$
$dv = sin(6x)/6 dx$
$v= -cos(6x)$

so $I = x^2(x/2-sin(6x)/12) -x^3/3 - x cos(6x) + sin(6x)/6$

How do you integrate int x^2 cos^2 3 x dx int x^2 cos^2 3 x dx using integration by parts?