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

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Answer 1 · 2016-01-07T02:15:39

$I = -x^2cos(x) + 2xsin(x) + 2cos(x) + c$

Say $u = x^2$ so $du = 2x$ , $dv = sin(x)$ so $v = -cos(x)$

$I = -x^2cos(x) +2intxcos(x)dx$

Say $u = x$ so $du = 1$ , $dv = cos(x)$ so $v = sin(x)$

$I = -x^2cos(x) + 2xsin(x) - 2intsin(x)dx$
$I = -x^2cos(x) + 2xsin(x) + 2cos(x) + c$

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