Question

How do you find the definite integral `int_(pi/2)^((5pi)/2) x^2 cos (1/5x)dx` ?

Answer 1

Find the indefinite integral
Subtract the indefinite integral evaluated at the lower limit from the indefinite integral evaluated at the upper limit.

Explanation:

Given: `int_(pi/2)^((5pi)/2) x^2 cos (1/5x)dx =`?

Find the indefinite integral:

`int x^2 cos (1/5x)dx`

Use integration by parts:

`intudv=uv-intvdu`

let `u = x^2` and `dv = cos(1/5x)`

Then `du = 2xdx` and `v = 5sin(1/5x)`

Substitute into the formula:

`int x^2 cos (1/5x)dx = 5x^2sin(1/5x)-10intxsin(1/5x)dx" "[1]`

Use integration by parts, again:

`intudv=uv-intvdu`

let `u = 10x` and `dv = -sin(1/5x)`

Then `du = 10dx` and `v = 5cos(1/5x)`

Substitute into the formula:

`-10intxsin(1/5x)dx = 50xcos(1/5x) - 50intcos(1/5x)dx" "[2]`

Substitute equation [2] into equation [1]:

`int x^2 cos (1/5x)dx = 5x^2sin(1/5x)+50xcos(1/5x) - 50intcos(1/5x)dx`

The last integral is trivial:

`int x^2 cos (1/5x)dx = 5x^2sin(1/5x)+50xcos(1/5x) - 250sin(1/5x)`

Please observe that I have not added a constant of integration, because we are going to use the results to evaluate a definite integral:

`int_(pi/2)^((5pi)/2) x^2 cos (1/5x)dx = 5((5pi)/2)^2sin(1/5(5pi)/2)+50((5pi)/2)cos(1/5(5pi)/2) - 250sin(1/5(5pi)/2) - (5(pi/2)^2sin(1/5pi/2)+50(pi/2)cos(1/5pi/2) - 250sin(1/5pi/2))`

`int_(pi/2)^((5pi)/2) x^2 cos (1/5x)dx ~~ 57.171`