How do you integrate $int x^2 cos3 x dx$ using integration by parts?

Question

3105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-09T01:49:51

$I = (9x^2sin(3x) +18xcos(3x) -2sin(3x))/9 + c$

$I = intx^2cos(3x)dx$

Say $u = x^2$ so $du = 2x$ and $dv = cos(3x)$ so $v = sin(3x)/3$

$I = (x^2sin(3x))/3 - 2/3intxsin(3x)dx$

Say $u = x$ so $du = 1$ and $dv = sin(3x)$ so $v = -cos(3x)/3$

$I = (x^2sin(3x))/3 - 2/3(-xcos(3x) +1/3intcos(3x)dx)$

$I = (x^2sin(3x))/3 - 2/3(-xcos(3x) +sin(3x)/9) + c$

$I = (9x^2sin(3x) +18xcos(3x) -2sin(3x))/9 + c$

How do you integrate int x^2 cos3 x dx int x^2 cos3 x dx using integration by parts?