Question #204957
Solve the integral of x²cosxdx, using integration by parts.
1
Expert's answer
2021-06-09T16:37:10-0400

x2cos(x)dx=x2sin(x)sin(x).2xdx=x2.sin(x)2x.sin(x)dx\int x^2cos(x)dx=x^2sin(x)- \int sin(x).2xdx=x^2.sin(x)-2\int x.sin(x)dx

u=x2   du=2x dxu=x^2 \space \space \space du=2x \space dx

dv=cos(x)dx   v=sin(x)dv=cos(x)dx \space \space \space v=sin(x)

x2sin(x)2(x.(cos(x)cos(x)dx)=x2sin(x)2(xcos(x)+cos(x)dxx^2sin(x)-2(x.(-cos(x)- \int-cos(x)dx)=x^2sin(x)-2(-xcos(x)+ \int cos(x)dx

u=x    du=dxu=x \space \space \space \space du=dx

dv=sin(x)dx     v=cos(x)dv=sin(x)dx \space \space \space \space \space v=-cos(x)

=x2.sin(x)+2xcos(x)2sin(x)+C=x^2.sin(x)+2xcos(x)-2sin(x)+C


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