How do you integrate int x^2 sin x dx x2sinxdx using integration by parts?

1 Answer
Jan 7, 2016

I = -x^2cos(x) + 2xsin(x) + 2cos(x) + cI=x2cos(x)+2xsin(x)+2cos(x)+c

Explanation:

Say u = x^2u=x2 so du = 2xdu=2x, dv = sin(x)dv=sin(x) so v = -cos(x)v=cos(x)

I = -x^2cos(x) +2intxcos(x)dxI=x2cos(x)+2xcos(x)dx

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

I = -x^2cos(x) + 2xsin(x) - 2intsin(x)dxI=x2cos(x)+2xsin(x)2sin(x)dx
I = -x^2cos(x) + 2xsin(x) + 2cos(x) + cI=x2cos(x)+2xsin(x)+2cos(x)+c