`int_(-pi/3)^(pi/3) x^4 sin(x) dx` Evaluate the definite integral.

You need to solve the definite integral, using fundamental theorem of calculus, such that:

`int_a^b f(x) dx = F(b) - F(a)`

First, you need to solve the indefinite integral `int x^4*sin x dx` , using parts, such that:

`int vdu = v*u - int udv`

If `v = x^4 and du = sin x` yields:

`dv = 4x^3 and u = -cos x`

`int x^4*sin x dx = -x^4*cos x + 4int x^3*cos x dx`

...

You need to solve the definite integral, using fundamental theorem of calculus, such that:

`int_a^b f(x) dx = F(b) - F(a)`

First, you need to solve the indefinite integral `int x^4*sin x dx` , using parts, such that:

`int vdu = v*u - int udv`

If `v = x^4 and du = sin x` yields:

`dv = 4x^3 and u = -cos x`

`int x^4*sin x dx = -x^4*cos x + 4int x^3*cos x dx`

Using parts again for `int x^3*cos x dx` yields:

`v = x^3 and du = cosx=> dv = 3x^2 and u = sin x`

`int x^3*cos x dx = x^3*sin x - 3int x^2*sin x dx`

Using parts again for `int x^2*sin x dx` yields:

`v = x^2 and du = sin x=> dv = 2x and u = -cos x`

`int x^2*sin x dx = -x^2*cos x + 2int x*cos x dx`

Using parts for the last time for `int x*cos x dx` yields:

`v = x and du = cos x => dv = 1 and u = sin x`

`int x*cos x dx = x*sin x - int sin x dx`

`int x*cos x dx = x*sin x + cos x`

Take the result `x*sin x + cos x` and replace it for `int x*cos x dx` above:

`int x^2*sin x dx = -x^2*cos x + 2x*sin x + 2cos x`

Take the result `-x^2*cos x + 2x*sin x + 2cos x` and replace it for `int x^2*sin x dx` above:

`int x^3*cos x dx = x^3*sin x - 3(-x^2*cos x + 2x*sin x + 2cos x)`

`int x^3*cos x dx = x^3*sin x + 3x^2*cos x - 6x*sin x - 6cos x`

Take the result `x^3*sin x + 3x^2*cos x - 6x*sin x - 6cos x` and replace it for `int x^3*cos x dx` above:

`int x^4*sin x dx = -x^4*cos x + 4int x^3*cos x dx`

`int x^4*sin x dx = -x^4*cos x + 4(x^3*sin x + 3x^2*cos x - 6x*sin x - 6cos x)`

`int x^4*sin x dx = -x^4*cos x + 4x^3*sin x + 12x^2*cos x - 24x*sin x - 24cos x`

Calculating the integral yields:

`int_(-pi/3)^(pi/3) x^4*sin x dx = (-x^4*cos x + 4x^3*sin x + 12x^2*cos x - 24x*sin x - 24cos x)|_(-pi/3)^(pi/3)`

`int_(-pi/3)^(pi/3) x^4*sin x dx = (-(pi/3)^4*cos (pi/3) + 4(pi/3)^3*sin (pi/3) + 12(pi/3)^2*cos (pi/3) - 24(pi/3)*sin (pi/3) - 24cos (pi/3) +(pi/3)^4*cos (pi/3) - 4(pi/3)^3*sin (pi/3) - 12(pi/3)^2*cos (pi/3) + 24(pi/3)*sin (pi/3) + 24cos (pi/3))`

Reducing like terms yields:

`int_(-pi/3)^(pi/3) x^4*sin x dx = 0`

You also may solve the integral by noticing the the function `f(x) = x^4*sin x` is odd and you may use the propert` int_(-a)^a f(x) dx = 0` if f(x) is odd.

You may prove that ` f(x) = x^4*sin x` is odd such that:

`f(-x) = (-x)^4*sin (-x) = x^4*(-sin x) = -x^4*sin x = -f(x)`

Hence, evaluating the definite integral, using either the fundamental theorem of calculus, or the property of odd functions `int_(-a)^a f(x) dx = 0` , yields `int_(-pi/3)^(pi/3) x^4*sin x dx = 0.`