Solve the integral of x²cosxdx, using integration by parts.
∫x2cos(x)dx=x2sin(x)−∫sin(x).2xdx=x2.sin(x)−2∫x.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∫x2cos(x)dx=x2sin(x)−∫sin(x).2xdx=x2.sin(x)−2∫x.sin(x)dx
u=x2 du=2x dxu=x^2 \space \space \space du=2x \space dxu=x2 du=2x dx
dv=cos(x)dx v=sin(x)dv=cos(x)dx \space \space \space v=sin(x)dv=cos(x)dx 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)dxx2sin(x)−2(x.(−cos(x)−∫−cos(x)dx)=x2sin(x)−2(−xcos(x)+∫cos(x)dx
u=x du=dxu=x \space \space \space \space du=dxu=x du=dx
dv=sin(x)dx v=−cos(x)dv=sin(x)dx \space \space \space \space \space v=-cos(x)dv=sin(x)dx v=−cos(x)
=x2.sin(x)+2xcos(x)−2sin(x)+C=x^2.sin(x)+2xcos(x)-2sin(x)+C=x2.sin(x)+2xcos(x)−2sin(x)+C
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