How to Integrate dz/ (y^2 + (x-z)^2))^1/2 Using Inverse Hyperbolic Functions?

[Arthy]

Integral of dz/ (y^2 + (x-z)^2))^1/2

 

[planck42]

Let x-z equal y sinh t

 

[HallsofIvy]

I would do it as a sequence of substitutions (planck42's use of a single substitution is correct and faster but harder to see).

First, because that "$(x- z)^2$" is a nuisance, let u= x- z. Then du= -dz so the integral becomes
$$-\int \frac{du}{y^2+ u^2}$$.

Now, remembering that $sin^2(\theta)+ cos^2(\theta)= 1$ so that $tan^1(\theta)+ 1= sec^2(\theta)$ and that $y^2+ u^2= y^2(1+ (u/y)^2$, let $u/y= tan(\theta)$ so that $du= y^2 sec^2(\theta)d\theta$ and $y^2+ u^2= y^2(1+ (u/y)^2)= y^1(sec^2(\theta))$ (Again, Planck42's hyperbolic substitution works fine but I learned trig substitutions before hyperbolic substutions so I tend to think of them first!). That makes the integral
$$\int\frac{y^2 sec^2(\theta)d\theta}{y^2 sec^2(\theta)}$$

 

[Arthy]

Thank you for the reply, but there is a square root at the denominator (y^2+ (x-z)^2)^1/2.

I know the answer of this derivation, with respect to limits 0 to L,

Integral of
dz/ Square root of (y^2+(x-z)^2) =
ln (x+ Square root of (x^2+y^2)/ x-L + square root of ((x-L)^2+y^2))

Please explain me the derivation part.

 

[Bohrok]

Going off of Halls' steps,
u/y = tanθ → u = y·tanθ
du = y·sec²θ dθ
$$\sqrt{(y^2+ u^2)} = \sqrt{(y^2(1+ (u/y)^2))} = \sqrt{(y^2(\sec^2\theta)} = y\sec \theta$$

Then you get
$$\int \frac{y \sec^2 \theta ~d\theta}{y\sec \theta} = \int \sec\theta ~d\theta$$

 

[planck42]

Let x-z be y sinh t
dz=-ycosh t dt

Using the identity $$cosh^{2}t=1+sinh^{2}t$$, we get

$${\int}-dt$$

Which is just -t. Now to put back the x's, y's, and z's.

$$x-z=ysinht$$ so
$$t=sinh^{-1}\frac{x-z}{y}$$

The rest is just a matter of inserting the bounds L and 0(and looking up the inverse hyperbolic functions; they're why your answer is a nasty ln function)