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

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Answer 1 · 2016-09-24T10:28:39

$1/2(sin(x^2)-x^2cos(x^2))+C$

$d/dx(x^2cos(x^2))=2xcos(x^2)-2x^3sin(x^2)$

but $2xcos(x^2)= d/dx(sin(x^2))$ so

$d/dx(x^2cos(x^2)) = d/dx(sin(x^2))-2x^3sin(x^2)$

then

$int x^3sin(x^2)dx = 1/2(sin(x^2)-x^2cos(x^2))+C$

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