Evaluate ` int x sin 3x dx`.

Explanation

Here, both the functions, viz., `x` and `sin 3x` are easily integrable and the derivative of x is one, a less complicated function.
Therefore, we take x as the first function and `sin 3x` as the second function. Thus,
`int underset (I)(x)underset(II)(sin)3x dx`
`=x{intsin 3xdx}-int{(d)/(dx)(x)intsin 3x dx}dx`
`= -x (cos 3x)/(3)-int1 {-(cos 3x)/(3)}dx`
`= -(1)/(3)x cos 3x +(1)/(3) int cos3x dx`
`= -(1)/(3)x cos 3x +(1)/(9) sin 3x +C`