I am trying to solve this integral, which is incorrect compared to Wolfram|Alpha. Why doesn't my method work?

Find $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$

Side work:

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$\sin{\theta} = \frac{3}{x}$

$x = \frac{3}{\sin{\theta}} = 3 \csc{\theta}$

$dx = -3 \csc{\theta}\cot{\theta} \ d\theta$

$-\int \frac{\sqrt{9\csc^2{\theta} - 9}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\tan{\theta} \ d\theta$

$-\int \frac{3 \cdot \sqrt{\csc^2{\theta} - 1}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\cot{\theta} \ d\theta$

$-\int \frac{\cot{\theta}\csc{\theta}\cot{\theta}}{3 \csc^3{\theta}} \ d\theta$

$-\frac{1}{3} \int \cos^2{\theta} \ d\theta$

$-\frac{1}{3} \int \frac{1}{2} \left(1 + \cos{\left(2\theta\right)}\right) \ d\theta$

$-\frac{1}{6} \int 1 + \cos{\left(2\theta\right)} \ d\theta$

$-\frac{1}{6} \int \ d\theta + \frac{1}{6} \int \cos{\left(2\theta\right)} \ d\theta$


$u = 2\theta$

$du = 2 \ d\theta$

$-\frac{1}{6} \int \ d\theta + \frac{1}{3} \int \cos{u} \ du$

$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{u} + C$

$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{2\theta} + C$


$\sin{\left(2\theta\right)} = 2 \sin{\theta} \cos{\theta}$

$\sin{\theta} = \frac{3}{x}$

$\cos{\theta} = \frac{\sqrt{x^2 - 9}}{x}$

$\theta = \sin^{-1}{\left(\frac{3}{x}\right)}$

$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2}{3} \cdot \frac{3}{x} \cdot \frac{\sqrt{x^2 - 9}}{x} + C$

$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2\sqrt{x^2 - 9}}{x^2} + C$

Thank you for your time.


The derivative of $\csc \theta$ is not $\csc \theta \tan \theta$ it is $-\csc \theta \cot \theta$.

After your edit:

Try writing $\sin^{-1} (3/x)$ in the form of $\tan^{-1}$.

Also you have the integral of $\cos 2\theta$ wrong.

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  • $\begingroup$ (cold chills) ... yes, your right, let me try re-evaluating $\endgroup$ – Oliver Spryn Mar 12 '12 at 22:27
  • $\begingroup$ Fixed the above work. Interestingly, it only resulted in a few sign changes. Any further ideas? $\endgroup$ – Oliver Spryn Mar 12 '12 at 22:41
  • $\begingroup$ @spryno724: Try writing $\sin^{-1} (3/x)$ in the form of $\tan^{-1}$... Also you have the integral of $\cos 2\theta$ wrong. $\endgroup$ – Aryabhata Mar 12 '12 at 22:51

Added: It has been pointed out that there may be some ambiguity about the integral the OP is trying to find. The introduction has an $x^2$ at the bottom, but in the solution the OP seems to be using $x^3$. That is much easier. The substitution $x=3\sec\theta$ transforms the integrand into a constant times $\frac{\tan^2\theta}{\sec^2\theta}$, which is just $\sin^2\theta$.

Original post: If we are going to use a trigonometric substitution, I would let $x=3\sec\theta$. (Secant is a bit more familiar than cosecant. You are therefore less likely to differentiate $\sec\theta$ incorrectly; incorrect differentiation of $\csc\theta$ spoiled our integration.)

Then $dx=3\sec\theta\tan\theta\,d\theta$. Plug in. From now on, I will forget about constants. So everything is a little wrong, but only by a constant factor.

On top we get $(\tan\theta)(\sec\theta\tan\theta)$. On the bottom we get $\sec^2\theta$. So we are interested in $$\frac{\tan^2\theta\sec\theta}{\sec^2\theta}.$$ Put everything in terms of sines and cosines. We get $\dfrac{\sin^2\theta}{\cos\theta}$.

There is a trick that one can use when there is an odd power of sine or cosine. In this case, multiply top and bottom by $\cos\theta$, and replace the $\cos^2\theta$ now at the bottom by $1-\sin^2\theta$. We have arrived at $$\int \frac{\cos\theta \,\sin^2\theta}{1-\sin^2\theta}\,d\theta.$$ Make the substitution $u=\sin\theta$. We arrive at $$\int \frac{u^2}{1-u^2}\,du.$$ The integrand is a rational function. It can be rewritten as $$\frac{1}{1-u^2}-1.$$ But (partial fractions) we find that $\dfrac{1}{1-u^2}=\dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1-u}\right)$.

Remark: As usual, there are many other approaches. For example, we can change $\frac{\tan^2\theta}{\sec\theta}$ to $\frac{\sec^2\theta-1}{\sec\theta}$, and then to $\sec\theta -\cos\theta$. If we happen to know the integral of $\sec\theta$, we are essentially finished. More strikingly, if we guess somehow that $F(x)$ is an antiderivative, then the magic substitution $u=F(x)$ solves the problem.

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  • $\begingroup$ He needs $\int\frac{\sqrt{x^2-9}}{x^3}dx$ not $\int\frac{\sqrt{x^2-9}}{x^2}dx$, though he put the latter one in the first paragraph (he solves with the cube). $\endgroup$ – Vadim Mar 13 '12 at 5:32

Let $I = \int \frac{\sqrt{x^2 – 9}}{x^2}$

Easiest is a hyperbolic substitution:

Put $x = 3\cosh{u}, u \le 0$, so $dx = 3\sinh(u)du$

Then \begin{align} I &=\int\frac{3\sinh(u^2)}{9 \cosh^2(u)}du\\\\ &= \int\tanh^2(u)du\\\\ &= \int\left(1–\mathrm{sech}^2(u)\right)du\\\\ &= u–\tanh(u)+c\\\\ &= u-\frac{3\sinh(u)}{3\cosh(u)}+c\\\\ &= \cosh^{-1}\left(\frac{x}{3}\right)-\frac{\sqrt{x^2–9}}{x}+c\\ \end{align}

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