Question

How do you integrate `int 1/sqrt(2-5x^2)` by trigonometric substitution?

Answer 1

`int \ 1/sqrt(2-5x^2) \ dx = 1/sqrt(5) \ arcsin (sqrt(5/2)x) + C`

Explanation:

We seek:

`I = int \ 1/sqrt(2-5x^2) \ dx`

Which, we can write as:

`I = int \ 1/sqrt(2(1-5/2x^2)) \ dx`
`\ \ = int \ 1/(sqrt(2)sqrt(1-(sqrt(5)/sqrt(2)x)^2)) \ dx`
`\ \ = 1/sqrt(2) \ int \ 1/(sqrt(1-(sqrt(5/2)x)^2)) \ dx`

We can now perform a substitution, Let:

`u = sqrt(5/2)x => (du)/dx = sqrt(5/2)`
`:. sqrt(2/5) (du)/dx = 1`

Substituting into the integral, we get:

`I = 1/sqrt(2) \ int \ 1/(sqrt(1-u^2)) \ (sqrt(2/5)) \ du`
`\ \ = 1/sqrt(5) \ int \ 1/(sqrt(1-u^2)) \ du`

Which is a standard integral, so we have:

`I = 1/sqrt(5) \ arcsin u + C`

Restoring the substitution:

`I = 1/sqrt(5) \ arcsin (sqrt(5/2)x) + C`