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)