# How to calculate the definite integral of the function x*cos3x if 0=<x=<pi/2?

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### 2 Answers

We have to find the integral of x*cos 3x.

This can be done using integration by parts.

Int [u dv] = u*v - Int[ v du]

Int[ x*cos 3x dx]

let u = x , du = dx

cos 3x dx = v => v = sin 3x / 3

=> x*sin 3x / 3 - Int[ (sin 3x / 3) dx]

=> x*sin 3x / 3 + cos 3x /9 + C

Between x = 0 an x = pi/2, the definite integral is:

(1/3)*pi/2 *sin 3*pi/2 + (1/9)cos 3*pi/2 - (1/9)cos 0

=> -pi/6 + 0 - (1/9)

=> -1/9 - pi/6

**The required definite integral is -1/9 - pi/6**

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.**