What is the integral of `xcos(x)`?

You use idea of the integrating by parts:
`int uv'dx = uv - intu'vdx`

`intx cosxdx =`
Let:
`u = x`
`u' = 1`
`v' = cosx`
`v = sinx`

Then:
`intx cosxdx = xsinx - int 1*sinxdx = xsinx - (-cosx) = xsinx+cosx`

The integral is:
`x*sin(x)+cos(x)+C`
You can get this result Integrating by Parts .
In general if you have the product of two functions `f(x)*g(x)` you can try this method in which you have:
`intf(x)*g(x)dx=F(x)*g(x)-intF(x)*g'(x)dx`

The integral of the product of the two functions is equal to the product of the integral ( `F(x)` ) of the first times the second function ( `g(x)` ) minus the integral of ther product of the integral of the first function ( `F(x)` ) times the derivative of the second function ( `g'(x)` ). Hopefully the last integral should be easier to solve than the starting one!!!

In your case you get (you can choose which one is `f(x)` to help you to make the solution easier ):

`f(x)=cos(x)`

`g(x)=x`

`F(x)=sin(x)`

`g'(x)=1`

And finally:
`intx*cos(x)dx=x*sin(x)-int1*sin(x)dx=x*sin(x)+cos(x)+C`
You can now check your answer by deriving this result.