How to calculate the definite integral of the function x*cos3x if 0=

The definite integral of the function will be evaluated
with the help of Leibniz Newton formula.

Int x*cos 3x dx =
F(pi/2) - F(0) (x = 0 and x = pi/2 are the limits of
integration)

We'll solve the integral by
parts:

Int udv = u*v - Int vdu
(*)

Let u = x => du =
dx

Let dv = cos 3x dx => v = (sin
3x)/3

We'll apply the formula
(*):

Int x*cos 3x dx = x*(sin 3x)/3 - Int (sin
3x)dx/3

Int x*cos 3x dx = x*(sin 3x)/3 + (cos
3x)/9

But Int x*cos 3x dx = F(pi/2) -
F(0)

F(pi/2) = (pi/2)*(sin 3pi/2)/3 + (cos
3pi/2)/9

F(pi/2) = - pi/6 +
0

F(pi/2) = - pi/6

F(0) = 0 +
(cos 0)/9

F(0) = 1/9

But Int
x*cos 3x dx = - pi/6 - 1/9

The definite
integral of the function y=x*cos 3x is: Int x*cos 3x dx = - pi/6 -
1/9.