Question

How do you integrate `int (2-x)/sqrt(x^2-4)dx` using trigonometric substitution?

Answer 1

`int (2-x)/sqrt(x^2-4) \ dx = 2ln|x + sqrt(x^2-4)| - sqrt(x^2-4) + C`

Explanation:

We seek:

`I = int (2-x)/sqrt(x^2-4) \ dx`

I would not use trigonometric substitutions. Using linearity we can split this into two integrals:

`I = int (2)/sqrt(x^2-4) - (x)/sqrt(x^2-4) \ dx`

`\ \ = int (2)/sqrt(x^2-4) \ dx - int \ (x)/sqrt(x^2-4) \ dx`

Consider the first integral:

`I_1 = int (2)/sqrt(x^2-4) \ dx`

We can perform a substitution of the form:

`u = x/2 => (du)/dx = 1/2`

Substituting into the integral, we get:

`I_1 = int (2)/sqrt((2u)^2-4) \ (2) du`
`\ \ \ = int (1)/sqrt(4(u^2-1)) \ du`
`\ \ \ = 2 int (1)/sqrt(u^2-1) \ du`

Which is a standard integral, so we can readily integrate and restore the substitution:

`I_1 = 2 ln|u+sqrt(u^2-1)|`
`\ \ \ = 2 ln |x/2 + sqrt(x^2/4-1) |`
`\ \ \ = 2 ln |x/2 + sqrt(1/4(x^2-4)) |`
`\ \ \ = 2 ln |x/2 + 1/2sqrt(x^2-4) |`
`\ \ \ = 2 ln |1/2(x + sqrt(x^2-4)) |`
`\ \ \ = 2 ln (1/2) + 2ln|x + sqrt(x^2-4)|`

Now, we consider the second integral:

`I_2 = int \ (x)/sqrt(x^2-4) \ dx`

We can perform a substitution of the form:

`u = x^2-4 => (du)/dx = 2x`

Substituting into the integral, we get:

`I_2 = int \ (1)/sqrt(u) \ (1/2) \ du`
`\ \ \ = 1/2 \ int \ (1)/sqrt(u) \ du`

Which is a standard integral, so we can readily integrate and restore the substitution:

`I_2 = (1/2 sqrt(u))/(1/2)`
`\ \ \ = sqrt(u)`
`\ \ \ = sqrt(x^2-4)`

And Combining these results, and introducing a constant of integration, we get:

`I = 2 ln (1/2) + 2ln|x + sqrt(x^2-4)| - sqrt(x^2-4) + C_1`
`\ \ = 2ln|x + sqrt(x^2-4)| - sqrt(x^2-4) + C`