We have to calculate the following integral quite often with Gauss surfaces in my Physics course. The teacher never exlained how we can find it and always gives us the result . How can I evaluate it?
$$\int \frac{dx}{(a^2+x^2)^{3/2}}$$ a is a constant
P.S. In case it's been uploaded I'll delete the question. I couldn't find it after giving a look though.
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In the first place, "solve" is the wrong word. You are evaluating an integral. One evaluates expressions; one solves equations; one solves problems.
If you see $(a^2+x^2)^{3/2}$ and don't think of the trigonometric substitution $x = a \tan\theta$ then you need to review trigonometric substitutions. Then you have $a^2+x^2 = a^2 (1 + \tan^2\theta) = a^2 \sec^2\theta,$ so $(a^2+x^2)^{3/2} = a^3 \sec^3\theta,$ and $dx = a\sec\theta\tan\theta\,d\theta.$ So you have $$ \int \frac{dx}{(a^2+x^2)^{3/2}} = \int \frac{a\sec^2\theta\,d\theta}{a^3\sec^3\theta} = \frac 1 {a^2} \int \cos\theta\,d\theta = \cdots\cdots. $$ Once you get a function of $\theta$ you need to convert it back into a function of $x$, and that will also require remembering some trigonometry.
If you make the substitution $x=a\tan\theta$, then $a^2+x^2=a^2\sec^2\theta$ and $\frac{dx}{d\theta}=a\sec^2\theta$, so the result is $$ \int\frac{a\sec^2\theta}{a^3\sec^3\theta}\;d\theta=a^{-2}\int \cos\theta\;d\theta=a^{-2}\sin\theta+C=\frac{x}{a^2\sqrt{a^2+x^2}}+C$$ Note that $\sin\theta$ can be determined using $\tan\theta=\frac{x}{a}$ and the Pythagorean theorem.