## How to compute this integral ?

As the title obviously states, how can I evaluate this integral by hand ? I know the result of it, I need to learn how to do it.

∫(z2+x2)-3/2dx

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 Recognitions: Gold Member Homework Help Science Advisor Letting "x" be the variable to integrate first, set x=z*Sinh(u), where Sinh(u) is the hyperbolic sine function.

 Quote by arildno Letting "x" be the variable to integrate first, set x=z*Sinh(u), where Sinh(u) is the hyperbolic sine function.
I didn't get it, it's from a simple example from a textbook. There should be a simpler way of doing this because it just skips the evaluation of this integral and directly passes to the result as if it's "that" easy to do. Any simpler solutions ?

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## How to compute this integral ?

 Quote by y33t I didn't get it, it's from a simple example from a textbook. There should be a simpler way of doing this because it just skips the evaluation of this integral and directly passes to the result as if it's "that" easy to do. Any simpler solutions ?
No.

it is NOT a particularly easy integral to evaluate, in that it is rather lengthy to do so.
That is probably why your book skipped it.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor x=z tan t can also do the trick.

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 Recognitions: Gold Member Homework Help Science Advisor The actual derivation is rather lengthy, but here it is: 1. x=zSinh(t). Thus, we have: $$dx=z\cosh(t)dt$$ $$z^{2}+x^{2}=z^{2}(1+\sinh^{2}(t))=z^{2}\cosh^{2}(t)$$ Thus, the integral can be simplified to: $$\int\frac{dt}{z|z|\cosh^{2}(t)}$$ 2. This is readily integrated to: $$\frac{\tanh(t)}{z|z|}=\frac{x}{z^{2}\sqrt{z^{2}+x^{2}}}$$
 Recognitions: Gold Member Science Advisor Staff Emeritus Arildno really likes the hyperbolic function substitutions. Personally, I prefer trig substitutions, perhaps only because they were the first ones I learned. We know, of course, that $sin^2(\theta)+ cos^2(\theta)= 1$ and, dividing through by $cos^2(\theta)$, $tan^2(\theta)+ 1= sec^2(\theta)$. So if we let $x= z tan(\theta)$, $z^2+ x^2= z^2+ z^2tan^2(\theta)= z^2(1+ tan^2(\theta)_= z^2sec^2(\theta)$. Of course, $dx= z sec^2(\theta)d\theta$ so the integral becomes $$\int\frac{z sec^2(\theta)}{z^3 sec^3(\theta)}d\theta= \int \frac{1}{sec(\theta)}d\theta$$ $$= \int cos(\theta) d\theta$$ which is easy. Since $\theta= arctan(x/z)$, the integral will eventually give $sin(arctan(x/z))$. You can imagine that as describing a right triangle with legs x and z (x opposite the angle) so that the hypotenuse has length $\sqrt{x^2+ z^2}$ and $sin(arctan(x/z))= \frac{x}{\sqrt{x^2+ z^2}}$.