Integrate $\frac{dy}{\sqrt{x^2+y^2}}$

Question

The problem is to evaluate the following integral: $$ \int \frac{dy}{\sqrt{x^2+y^2}} $$ If in the numerator was $y\cdot dy$, it would be absolutely simple (straightforward application of the substitution rule). But in this case, I don't know how to proceed. Any hints welcome.

Answer 1

Use the identity $$ \frac{d}{ds} \sinh^{-1} s = \frac{1}{\sqrt{1 + s^2}} $$ and change of variable $y \mapsto y/x, dy\mapsto x^{-1}\,dy$.

Note: Although you will get an indefinite integral in terms of $\sinh^{-1}$, you can convert that to logarithmic form.

Answer 2

HINT:

Put $y=x\tan \theta$ and substitute.
Then $dy=x\sec^2 \theta d\theta$.

So the integral becomes $$\int \frac{dy}{\sqrt{x^2+y^2}}=\int \frac{x\sec^2 \theta d\theta}{\sqrt{x^2+x^2\tan^2 \theta}}=\int \frac{x\sec^2 \theta d\theta}{x\sec \theta}$$

Answer 3

The function $\ln(y+\sqrt{y^2+x^2})$ has a "magical" property. By a nice simplification,

$$\left(\ln(y+\sqrt{y^2+x^2})\right)'_y=\frac{1+\dfrac y{\sqrt{y^2+x^2}}}{y+\sqrt{y^2+x^2}}=\frac1{\sqrt{y^2+x^2}}.$$

But you can retrieve this result by the substitution $y=x\sinh(t)$, so that $\sqrt{y^2+x^2}=x\cosh(t)$.