# Integration trig substitution $\int \frac{dx}{x\sqrt{x^2 + 16}}$

$$\int \frac{dx}{x\sqrt{x^2 + 16}}$$

With some magic I get down to $$\frac{1}{4} \int\frac{1}{\sin\theta} d\theta$$

Now is where I am lost. How do I do this? I tried integration by parts but it doesn't work.

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–  1015 Jul 2 '13 at 21:29
I wonder how you got to $\frac{d\theta}{\sin\theta}$. The obvious substitution I see leads to $\frac{dt}{\cosh t}$. –  Daniel Fischer Jul 2 '13 at 21:36
How is a hyperbolic subsitution ever obvious? Lol –  Dantheman Jul 2 '13 at 21:39
@DanielFischer Yes, another standard way of dealing with that is the trig $x=4\sec\theta$ which yields $\int \frac{4\sec\theta \tan\theta d\theta }{4\sec\theta 4\tan\theta }=\int \frac{d\theta}{4}=\frac{\mbox{arcsec} (x/4)}{4}+C$. If it really is $-16$, the OP made a mistake. If is $+16$, then $x=4\tan\theta$ yields cosec indeed. –  1015 Jul 2 '13 at 21:43
What was the magic substitution from $$\int \frac{dx}{x\sqrt{x^{2}-16}}$$ to $$\frac{1}{4} \int\frac{1}{\sin\theta} d\theta\ ?$$ –  Américo Tavares Jul 2 '13 at 23:34

HINT:

$$\int\frac{d\theta}{\sin\theta}=\int\csc\theta d\theta=\int\frac{\csc\theta(\csc\theta+\cot\theta)}{\csc\theta+\cot\theta}d\theta$$

Now what’s the derivative of that last denominator?

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You seemed to have massively complicated things. Also I see what you did but how would I never to do that? –  Dantheman Jul 2 '13 at 21:31
@Dantheman Not really. That is the standard workaround such integral. –  Pedro Tamaroff Jul 2 '13 at 21:35
@Dantheman: It’s a standard formula and a standard derivation of it. I find it a little easier to remember the idea behind the derivation than to remember the antiderivative itself, especially since a very similar trick takes care of $\int\sec\theta d\theta$. For these two integrals you have to remember something, and I prefer ideas to formulas. –  Brian M. Scott Jul 2 '13 at 21:36
So I need to memorize that to find the integral of sec and cot I need to know to multiply by its derivative over its derivative? So I need to memorize 2 more derivatives and that the method to get these integrals is that? –  Dantheman Jul 2 '13 at 21:38
Also is scs + cot useful? It isn't a derivative of anything. –  Dantheman Jul 2 '13 at 21:38

UPDATE 2 The first integral is not $\int \frac{dx}{x\sqrt{x^{2}-16}}$ but $\int \frac{dx}{x\sqrt{x^{2}+16}}$ (as noticed by julien), because

\begin{eqnarray*} I &=&\int \frac{dx}{x\sqrt{x^{2}+16}},\qquad x=\tan \theta ,dx=\sec ^{2}\theta d\theta \\ &=&\int \frac{\sec ^{2}\theta }{\left( \tan \theta \right) 4\sec \theta }% \,d\theta =\int \frac{\sec \theta }{4\tan \theta }\,d\theta \\ &=&\int \frac{\sec \theta }{4\tan \theta }\,d\theta =\int \frac{1}{4\sin \theta }\,d\theta \text{.} \end{eqnarray*}

Use the Weierstrass substitution $$t=\tan\frac{\theta}{2}.$$

Then

$$\int \frac{1}{\sin \theta }\,d\theta =\int \frac{2}{\frac{2t}{1+t^{2}} \left( 1+t^{2}\right) }\,dt=\int \frac{1}{t}\,dt=\ln \left\vert t\right\vert +C=\ln \left\vert \tan \frac{\theta }{2}\right\vert +C.$$

Comment: The Weierstrass substitution is a universal standard substitution to evaluate an integral of a rational fraction in $\sin \theta,\cos \theta$, i.e. a rational fraction of the form

$$R(\sin \theta,\cos \theta)=\frac{P(\sin \theta,\cos \theta)}{Q(\sin \theta,\cos \theta)},$$

where $P,Q$ are polynomials in $\sin \theta,\cos \theta$

$$\begin{equation*} \tan \frac{\theta }{2}=t,\qquad\theta =2\arctan t,\qquad d\theta =\frac{2}{1+t^{2}}dt \end{equation*},$$

which converts the integrand into a rational function in $t$. We know from trigonometry that

$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{ \theta}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2} \frac{\theta }{2}}=\frac{2t}{1+t^2}.$$

Proof. A possible proof is the following one, which uses the double-angle formulas and the identity $\cos ^{2}\frac{\theta}{2}+\sin ^{2}\frac{\theta}{2}=1$:

$$\begin{eqnarray*} \cos \theta &=&\cos ^{2}\frac{\theta}{2}-\sin ^{2}\frac{\theta }{2}=\frac{\frac{\cos ^{2} \frac{\theta}{2}-\sin ^{2}\frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}{\frac{\cos ^{2} \frac{\theta}{2}+\sin ^{2}\frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}=\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2}\frac{\theta}{2}}, \\ && \\ \sin \theta &=&2\sin \frac{\theta}{2}\cos \frac{\theta}{2}=\frac{\frac{2\sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}{\frac{\cos ^{2}\frac{\theta}{2}+\sin ^{2} \frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}=\frac{2\tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}. \end{eqnarray*}$$

Another possible substitution is the Euler substitution $$\begin{equation*} \sqrt{x^{2}+16}=t+x. \end{equation*}$$ Then $$\begin{eqnarray*} I &=&\int \frac{dx}{x\sqrt{x^{2}+16}}=\int \frac{2}{t^{2}-16}\,dt \\ &=&\int \frac{1}{4\left( t-4\right) }-\frac{1}{4\left( t+4\right) }dt=\frac{1 }{4}\ln \left\vert \frac{t-4}{t+4}\right\vert +C \\ &=&\frac{1}{4}\ln \left\vert \frac{\sqrt{x^{2}+16}-x-4}{\sqrt{x^{2}+16}-x+4} \right\vert +C. \end{eqnarray*}$$

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Well that is disheartening. More stuff to memorize? –  Dantheman Jul 2 '13 at 21:40
@Dantheman You insist on memorizing. Try learning instead. Comprehending what's going it. It will benefit you the most. –  Pedro Tamaroff Jul 2 '13 at 21:41
So more stuff to learn? I already don't have enough time to learn what we do in class, how could I possibly learn a whole new method? –  Dantheman Jul 2 '13 at 21:42
@Dantheman No, this substitution is a universal substitution for a rational function in $\sin\theta,\cos\theta$. Sometimes there are faster methods but this substitution is worth knowing. It allows you to reduce the integral to one of a rational function in $t$. –  Américo Tavares Jul 2 '13 at 21:42
Believe me when I say this. The test material has been stupidly designed. Learn this method and you will be among the top students in integration. This method allows you to tackle any of the trigonometric integrals they will give you. In addition, the original integral can be solved by a substitution that is not trigonometric called Euler's substitutions. I will add an answer talking a little about it. They allow you to solve almost any of the integrals with radicals they will give you. –  ABC Jul 2 '13 at 21:50

Hint: \begin{align} \int\frac1{\sin(\theta)}\,\mathrm{d}\theta &=\int\frac{\sin(\theta)}{\sin^2(\theta)}\,\mathrm{d}\theta\\ &=-\int\frac1{1-\cos^2(\theta)}\,\mathrm{d}\cos(\theta)\\ &=-\frac12\int\left(\frac1{1-\cos(\theta)}+\frac1{1+\cos(\theta)}\right)\,\mathrm{d}\cos(\theta)\\ \end{align}

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What is dcos? I domt follow –  Dantheman Jul 3 '13 at 0:51
@Dantheman: $\mathrm{d}\cos(\theta)=-\sin(\theta)\,\mathrm{d}\theta$. You can think of this as the substitution $u=\cos(\theta)$ and then $$\int\frac1{\sin(\theta)}\,\mathrm{d}\theta =-\frac12\int\left(\frac1{1-u} + \frac1{1+u}\right) \,\mathrm{d}u$$ –  robjohn Jul 3 '13 at 6:52

I think the best answer to the integral of the trigonometric function is the one given by Americo Tavares. Now, the original integral can be treated directly without the unnecessary trigonometric substitution.

Often forgotten in current Calculus textbooks. Are the Euler's substitutions that allow you to solve not only that integral but every one of the form $\int R(\sqrt{ax^2+bx+c},x)\text{d}x$, where $R(x,y)$ is any rational function.

These substitutions transform your integral (and any of the form above) into the integral of a rational function. From there the problem is solved because we know how to compute integrals of any rational function.

PS: The algorithm I linked for computing integrals of rational functions is not the only one. There are many others.

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Let's see if another method works: $$\int \frac{dx}{x\sqrt{x^2+16}} = \int\frac{x\,dx}{x^2\sqrt{x^2+16}} = \int \frac{du/2}{(u-16)\sqrt{u}}$$ where $u=x^2+16$ so that $du=2x\,dx$. Now let $w=\sqrt{u}$ so that $w^2=u$ and $2w\,dw=du$. Then we have $$\int\frac{w\,dw}{(w^2-16)w} = \int\frac{dw}{(w-4)(w+4)} =\int\left(\frac{\bullet}{w-4} + \frac{\bullet}{w+4}\right) \, dw,\quad\text{ etc.}$$

Conclusion: It can be done without a trigonometric substitution.

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That seems much easier, trig is like magic to me and I don't have all the indentities memorized. I just fear that the test will be full of trick questions. –  Dantheman Jul 3 '13 at 1:23
@Dantheman : Glad you liked it. Trigonometric substitutions aren't really trick questions, but you do need to know some things to use them. –  Michael Hardy Jul 3 '13 at 2:57