Question

How do you integrate `int dx/sqrt(x^2+4)` using trig substitutions?

Answer 1

Substitute `x=2 sinh u` quickly getting the integral to be `sinh^{-1}(x/2)+C`

Explanation:

`x=2sinh u`, `dx=2cosh u du` giving the integral as
`int2coshu/ sqrt(4 sinh^2u+4)du`
`=int cosh u/ cosh u du` because `cosh^2u -sinh^2 u =1`
`=int1du`
`=u+C`
`=sinh^-1(x/2)+C`

Some people prefer to write this as `ln(x+sqrt(x^2+4))` which is equivalent, as it differs from the first solution by a fixed constant `ln(1/2)`.