How does one integrate $$\int \dfrac{du}{(a^2 + u^2)^{3/2}}\ ?$$

The table of integrals here: http://teachers.sduhsd.k12.ca.us/abrown/classes/CalculusC/IntegralTablesStewart.pdf

Gives it as: $$\frac{u}{a^2 ( a^2 + u^2)^{1/2}}\ .$$

I'm getting back into calculus and very rusty. I'd like to be comfortable with some of the proofs behind various fundamental "Table of Integrals" integrals.

Looking at it, the substitution rule seems like the method of choice. What is the strategy here for choosing a substitution? It has a form similar to many trigonometric integrals, but the final result seems to suggest that they're not necessary in this case.


A trigonometric substitution does indeed work.

We want to express $(a^2 + u^2)^{3/2}$ as something without square roots. We want to use some form of the Pythagorean trigonometric identity $\sin^2 x + \cos^2 x = 1$. Multiplying each side by $\frac{a^2}{\cos^2 x}$, we get $a^2 \tan^2 x + a^2 = a^2 \sec^2 x$, which is in the desired form. of (sum of two squares) = (something squared).

This suggests that we should use the substitution $u^2 = a^2 \tan^2 x$. Equivalently, we substitute $u = a \tan x$ and $du = a \sec^2 x dx$. Then $$ \int \frac{du}{(a^2 + u^2)^{3/2}} = \int \frac{a \sec^2 x \, dx}{(a^2 + a^2 \tan^2 x)^{3/2}}. $$ Applying the trigonometric identity considered above, this becomes $$ \int \frac{a \sec^2 x \, dx}{(a^2 \sec^2 x)^{3/2}} = \int \frac{dx}{a^2 \sec x} = \frac{1}{a^2} \int \cos x \, dx, $$ which can be easily integrated as $$ =\frac{1}{a^2} \sin x. $$ Since we set $u = a \tan x$, we substitute back $x = \tan^{-1} (\frac ua)$ to get that the answer is $$ =\frac{1}{a^2} \sin \tan^{-1} \frac{u}{a}. $$ Since $\sin \tan^{-1} z = \frac{z}{\sqrt{z^2 + 1}}$, this yields the desired result of $$ =\frac{u/a}{a^2 \sqrt{(u/a)^2 + 1}} = \frac{u}{a^2 (a^2 + u^2)^{1/2}}. $$

  • 6
    $\begingroup$ An even easier way to find $\sin(\tan^{-1}(\frac{u}{a}))$ is to draw a right triangle with legs $u$ and $a$, and hypotenuse $\sqrt{u^2+x^2}$ Then set $\theta$ such that $\tan^{-1}(\frac{u}{a}) = \theta$. Then simply find $\sin \theta$: $\sin \theta = \frac{u}{\sqrt{a^2+u^2}}$ $\endgroup$
    – daviewales
    Apr 7 '14 at 7:13

Let $u = a \tan \theta$ and work from there. See this (page about trig substitutions).

  • $\begingroup$ Link doesn't work $\endgroup$
    – zabop
    Dec 8 '19 at 17:13

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