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Let $A$ be a constant and $z$ a variable. Evaluate the integral: $$\int \frac{\mathrm dz}{(A^2+z^2)\sqrt{2A^2+z^2}}$$

Note: I've tried the most common trigonometric substitutions$($like $z = A\tan(\theta))$, but had no success.

Integreek
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user2288551
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    Just making a suggestion that using $z=A\tan \theta$ would not make it very easy because here the main cause of trouble is the root in denominator and the substitution does not help to get rid of it. Also another substitution you can use is putting $z=\frac 1x$ and then substitute the term inside root as $y$. – sonu Feb 22 '18 at 15:06

5 Answers5

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Let $z=\sqrt{2}A\tan \theta$

The integral becomes (I will just skip the trivial steps) $$\int \frac{\cos\theta d\theta}{A^2(1+\sin^2\theta)} = \frac{1}{A^2}\int \frac{ d\sin\theta}{1+\sin^2\theta} = \frac{\arctan(\sin\theta)}{A^2}$$

I think this is clear enough and you can figure out the rest.

user2288551
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user260822
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you Can use the Substitution $$\sqrt{2A^2+z^2}=t+z$$ then you will get $$z=\frac{2A^2-t^2}{2t}$$ and $$dz=-\frac{2 A^2+t^2}{2 t^2}dt$$

Dr. Sonnhard Graubner
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An efficient method that doesn’t require any trig substitution is the substitution $z=\frac1{t}$:

$$\begin{align}\int\frac{\mathrm dz}{(A^2+z^2)\sqrt{2A^2+z^2}}&=\frac{-\text{sgn}A}{A^3}\int\frac{t\mathrm dt}{\left(t^2+\frac1{A^2}\right)\sqrt{2t^2+\frac1{A^2}}}\\&\overset{\sqrt{2t^2+\frac1{A^2}}=u}{=} \frac{-\text{sgn}A}{4A^3}\int\frac{\mathrm du}{\frac{u^2}2+\frac1{2A^2}}\\&=\cdots\end{align}$$

Integreek
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You are on the right track in recognizing the use of the substitution $z=A\tan\theta$; it is indeed feasible:

$$\begin{align}\int\frac{\mathrm dz}{(A^2+z^2)\sqrt{2A^2+z^2}}&=\frac1{A}\int\frac{\mathrm d\theta}{\sqrt{A^2+A^2\sec^2\theta}}\\&=\frac1{A^2\text{sgn}A}\int\frac{\cos\theta\mathrm d\theta}{\sqrt{2-\sin^2\theta}}\\&=\frac1{A^2\text{sgn}A}\sin^{-1}\left(\frac{\sin\theta}{\sqrt2}\right)+C\end{align}$$

Now, just do the back-substitution.

Integreek
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Just like the trig substitution you thought of, an analogous hyperbolic substitution $z=\sqrt2|A|\sinh\theta$ can be performed since $\frac{\mathrm d}{\mathrm dx}\sinh^{-1}\left(\frac{x}{a}\right)=\frac1{\sqrt{x^2+a^2}}$:

$$\begin{align}\int\frac{\mathrm dz}{(A^2+z^2)\sqrt{2A^2+z^2}}&=\int\frac{\mathrm d\theta}{A^2(\cosh^2\theta+\sinh^2\theta)}\\&=\frac1{A^2}\int\text{sech}2\theta\mathrm d\theta\\&=\frac1{2A^2}\tan^{-1}(\sinh2\theta)+C\end{align}$$

Integreek
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