How do you evaluate the integral $int dx/sqrt(a^2+x^2)$ ?

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Answer 1 · 2017-02-24T18:40:23

$= sinh^(-1) (x/a) + C$

Because $cosh^2 z - sinh^2 z = 1$ , I would go with the hyperbolic sub $x = a sinh y, dx = a cosh y \ dy$

The integral becomes:

$int 1/sqrt(a^2+a^2 sinh^2 y)a cosh y \ dy$

$= int 1/(a cosh y)*a cosh y \ dy$

$= y + C$

$= sinh^(-1) (x/a) + C$

And because $sinh^(-1) z = ln ( z + sqrt(z^2 + 1))$

$= ln ( x/a + sqrt((x/a)^2 + 1)) + C$

How do you evaluate the integral int dx/sqrt(a^2+x^2)int dx/sqrt(a^2+x^2)?