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I'm trying to evaluate this integral: $$\int\frac{\mathrm dx}{(a^2x^2-b^2)^\frac32}$$

And while certain online solvers give me a certain trigonometric solution, I've tried a different one that I think should work. My own work's below:

$$=\int\frac{(a^2x^2+b^2)^\frac12}{(a^2x^2+b^2)^2}\mathrm dx$$

$$=\int\frac{\left(a^2\left(\frac{b\tan\theta}{a}\right)^2+b^2\right)^\frac12}{\left(a^2\left(\frac{b\tan\theta}{a}\right)^2+b^2\right)^2}\mathrm dx$$ $$=\int\frac{\left(\frac{a^2b^2\tan^2\theta}{a^2}+b^2\right)^\frac12}{\left(\frac{a^2b^2\tan^2\theta}{a^2} +b^2\right)^2}\mathrm dx$$ $$=\int\frac{(b^2\tan^2\theta+b^2)^\frac12}{(b^2\tan^2\theta+b^2) ^2}\mathrm dx$$ $$=\int\frac{\sqrt{b^2(\tan^2\theta+1)}}{(b^2(\tan^2\theta+1))^2}\mathrm dx$$ $$=\int\frac{\sqrt{b^2\sec^2\theta}}{(b^2\sec^2\theta)^2}$$ $$=\int\frac{b\sec\theta}{b^4\sec^4\theta}$$ $$=\int\frac1{b^3\sec^3\theta}$$ $$=\frac1{b^3}\int\frac1{\sec^3\theta}$$ $$=\frac1{b^3}\int\cos^3\theta$$ $$=\frac{\sin\theta}{b^3}-\frac{\sin^3\theta}{3b^3}$$

$$=\frac{3\sin\theta}{3b^3}-\frac{\sin^3\theta}{3b^3}$$ $$=\frac{\sin\theta(3-\sin^2\theta)}{3b^3}$$ $$=\frac{ax}{3b^3\sqrt{a^2x^2+b^2}}\left(3-\frac{a^2x^2}{a^2x^2+b^2}\right)$$

That said, I seem to not be getting the right answer. This is what integral-calculator.com gives me:

$$-\frac{x}{b^2\sqrt{a^2x^2-b^2}}+C$$

And having graphed both in Desmos, both the anti-derivatives look quite different when graphed for $a = b = 1$:

enter image description here

Could someone help me figure out what I'm missing here?

Integreek
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solar_microwave_32
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    you have $x=\frac{b}{a}\tan(\theta)$, which is fine, but you failed to replace $dx$ with $\frac{b}{a}\sec^2(\theta)d\theta$ in your solutions – C Squared Jan 31 '21 at 01:31
  • Welcome. LaTeX is a bit of a job to figure out but it does repay itself with dividends! You may generally find that questions that do not require clicking into a bunch of third-party links are answered more quickly. But I really appreciate the effort that has gone into wrestling with this problem. – leslie townes Jan 31 '21 at 01:37
  • Related: https://math.stackexchange.com/questions/3139536/integrate-int-fracdxx2-1-frac32-via-trig-substitution/5000717#5000717 – Integreek Nov 23 '24 at 06:38
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    Some points about terminology: An indefinite integral is an operation on a function, not an equation; hence it is evaluated, not solved. The function obtained on integration is called an anti-derivative, not an equation. See Wikipedia. – Integreek Nov 23 '24 at 11:31

1 Answers1

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You would be on the right track had the integral been

$$\int\frac{\mathrm dx}{(a^2x^2\color{red}{+}b^2)^\frac32}$$

Even if the integral was this, you have made a few mistakes:

$1.$ You have forgotten the arbitrary constant $+C$.

$2$. You have forgotten to write $\mathrm d\theta$ halfway through your solution, if you wrote it, you’d realize that after performing the substitution $x=\frac{b}{a}\tan\theta$, the differential $\mathrm dx$ also needs to be rewritten is terms of $\theta$ and $\mathrm d\theta$:

$$x=\frac{b}{a}\tan\theta\implies\mathrm dx=\frac{b}{a}\sec^2\theta\,\mathrm d\theta$$

$3.$ You have not specified the domain of $\theta$. This is important because:

  • The substitution $x=x(\theta)$ must be bijective so that there is no ambiguity in back-substitution.
  • This will have to be taken into consideration while taking square roots of even powers and back-substituting in the end.

For example, if you let $\theta\in\left(2n\pi-\frac\pi2,2n\pi+\frac\pi2\right)$ which is the most convenient domain in your integral as $\sec\theta>0\implies\sqrt{\sec^2\theta}=\sec\theta$, and the signs of $\sin\theta$ and $x$ are the same($\sin\theta$ is the anti-derivative of the given integral, see below), so we can write $$\sin\theta=\frac{ax}{\sqrt{a^2x^2+b^2}}, \cos\theta=\frac{b}{\sqrt{a^2x^2+b^2}}$$ but if $\theta\in[2n\pi, (2n+1)\pi)-\left\{2n\pi+\frac\pi2\right\}$, we need to bifurcate this into $2$ cases, $\theta\in\left[2n\pi, 2n\pi+\frac\pi2\right)$ and $\theta\in\left(2n\pi+\frac\pi2, (2n+1)\pi\right)$:

$$\theta\in\left[2n\pi, 2n\pi+\frac\pi2\right)\implies\sin\theta= \frac{ax}{\sqrt{a^2x^2+b^2}},\cos\theta=\frac{b}{\sqrt{a^2x^2+b^2}}$$

$$\theta\in\left(2n\pi+\frac\pi2, (2n+1)\pi\right) \implies\sin\theta= \frac{-ax}{\sqrt{a^2x^2+b^2}},\cos\theta=\frac{-b}{\sqrt{a^2x^2+b^2}}$$

which unnecessarily complicates things.

Also, there was no need to complicate the first part of your solution so much, you could’ve simply let the power in the denominator be $\frac32$, which would later get converted to $3$ due to $\sec^2\theta$.

Let me also share the correct solution for your reference:

$$\require{cancel}\begin{align}\int\frac{\mathrm dx}{(a^2x^2+b^2)^\frac32}&\overset{x=\frac{b}{a}\tan\theta, \text{ }\theta\in\left(-\frac\pi2,\frac\pi2\right)}{=}\int\frac{\frac{\cancel{b}}{a}\cancel{\sec^2\theta}\mathrm d\theta}{b^{\cancel{{3}}^{2}}\sec^\cancel{3}\theta}\\&=\frac1{ab^2}\int\cos\theta\,\mathrm d\theta\\&=\frac{\sin\theta}{ab^2}+C\\&=\frac{x}{b^2\sqrt{a^2x^2+b^2}}+C\end{align}$$

Now, coming back to your original integral

$$\int\frac{\mathrm dx}{(a^2x^2-b^2)^\frac32}$$

a possible method is to substitute $x=\frac{b}{a}\sec\theta, \theta\in(0,\pi)-\{\frac\pi2\}$ as we need $\tan\theta>0$ since we end up $\left(b^2\tan^2\theta\right)^\frac32$ upon performing the substitution. Another faster substitution is $x=\frac1v$ since this is a standard technique to integrate functions of the type $\dfrac1{(ax^2+b)\sqrt{cx^2+d}}$:

$$\begin{align}\int\frac{\mathrm dx}{(a^2x^2-b^2)^\frac32}&=-\operatorname{sgn}x\int\frac{v}{(a^2-b^2v^2)^\frac32}\mathrm dv\\&=\frac{\operatorname{sgn}x}{2b^2}\int\frac{\mathrm d(a^2-b^2v^2)}{(a^2-b^2v^2)^\frac32}\\&=\frac{-\operatorname{sgn}x}{b^2\sqrt v}+C\\&=\frac{-x}{b^2\sqrt{a^2x^2-b^2}}+C\end{align}$$

which is the same result obtained on replacing $b$ with $\pm ib$ in the integral of $(a^2x^2+b^2)^\frac{-3}2$.

Integreek
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