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

1 Answer

Sep 24, 2016

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

Explanation:

#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#

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