Question

How do you evaluate `int_1^sqrt(3) 4/(x^2sqrt(x^2 - 1)) dx`?

Answer 1

The integral has value `4sqrt(2/3)`

Explanation:

Use the substitution `x = sectheta`. Then `dx = secthetatantheta d theta`. You should (stylistically) change the bounds of integration, but for ease of entering on the computer, I haven't. Call the integral `I`.

`I = int_1^sqrt(3) 4/((sec theta)^2sqrt((sec theta)^2 - 1)) secthetatantheta d theta`

`I = int_1^sqrt(3) 4/(sec^2thetasqrt(tan^2theta)) secthetatantheta d theta`

`I = int_1^sqrt(3) 4/(sec^2thetatantheta) secthetatantheta d theta`

`I= int_1^sqrt(3) 4/sectheta d theta`

`I = int_1^sqrt(3) 4costheta d theta`

`I = [4sintheta]_1^sqrt(3)`

DO NOT EVALUATE THIS INTEGRAL. SINCE WE DIDN'T CHANGE THE BOUNDS OF INTEGRATION, THIS INTEGRAL WILL BE INCORRECT. THE CORRECT PROCESS WOULD BE TO SWITCH THE VARIABLE BACK TO `X`.

From our initial substitution, we know that `sectheta = x/1`, so if we were to draw a triangle, the hypotenuse would measure `x` and the side adjacent `theta` would measure `1`. By pythagoras, the side opposite would measure `sqrt(x^2 - 1)`. Therefore, `sintheta = sqrt(x^2- 1)/x`.

`I = [(4sqrt(x^2 - 1))/x]_1^sqrt(3)`

`I = (4sqrt(sqrt(3)^2 - 1))/sqrt(3) - (4sqrt(1^2 - 1))/1`

`I = (4sqrt(2))/sqrt(3) - 0`

`I = 4sqrt(2/3)`

Hopefully this helps!