I've got another problem with my CalcII homework. The problem deals with trig substitution in the integral for integrals following this pattern: $\sqrt{a^2 + x^2}$. So, here's the problem:

$$\int_{-2}^2 \frac{\mathrm{d}x}{4 + x^2}$$

I graphed the function and because of symmetry, I'm using the integral: $2\int_0^2 \frac{\mathrm{d}x}{4 + x^2}$

Since the denominator is not of the form: $\sqrt{a^2 + x^2}$ but is basically what I want, I ultimately decided to take the square root of the numerator and denominator:

$$2 \int_0^2 \frac{\sqrt{1}}{\sqrt{4+x^2}}\mathrm{d}x = 2 \int_0^2 \frac{\mathrm{d}x}{\sqrt{4+x^2}}$$

From there, I now have, using the following: $\tan\theta = \frac{x}{2} => x = 2\tan\theta => dx = 2\sec^2\theta d\theta$

$$ \begin{array}{rcl} 2\int_{0}^{2}\frac{\mathrm{d}x}{4+x^2}\mathrm{d}x & = & \sqrt{2}\int_{0}^{2}\frac{\mathrm{d}x}{\sqrt{4+x^2}}\mathrm{d}x \\ & = & \sqrt{2}\int_{0}^{2}\frac{2\sec^2(\theta)}{\sqrt{4+4\tan^2(\theta)}}\mathrm{d}\theta \\ & = & \sqrt{2}\int_{0}^{2}\frac{2\sec^2(\theta)}{2\sqrt{1+\tan^2(\theta)}}\mathrm{d}\theta \\ & = & \sqrt{2}\int_{0}^{2}\frac{\sec^2(\theta)}{\sqrt{\sec^2(\theta)}}\mathrm{d}\theta \\ & = & \sqrt{2}\int_{0}^{2}\frac{\sec^2(\theta)}{\sec(\theta)}\mathrm{d}\theta \\ & = & \sqrt{2}\int_{0}^{2}\sec(\theta)\mathrm{d}\theta \\ & = & \sqrt{2}\left [\ln{\sec(\theta)+\tan(\theta)} \right|_{0}^{2}] \\ & = & \sqrt{2}\left [ \ln{\frac{\sqrt{4+x^2}}{2}+\frac{x}{2} } \right|_{0}^{2} ] \end{array} $$

I'm not sure if I've correctly made the integral look like the pattern it's supposed to have. That is, trig substitutions are supposed to be for $\sqrt{a^2 + x^2}$ (in this case that is, there are others). This particular problem is an odd numbered problem and the answer is supposed to be $\frac{\pi}{4}$. I'm not getting that. So, the obvious question is, what am I doing wrong? Also note, I had trouble getting the absolute value bars to produce for the ln: don't know what I did wrong there either.

Thanks for any help, Andy

  • 1
    $\begingroup$ Why is it OK to take the square root of numerator and denominator? By your reasoning, $1/2 = 1/\sqrt{2}$. $\endgroup$ – Ron Gordon Apr 1 '13 at 20:43
  • $\begingroup$ You do know that $\displaystyle \int \frac{dx}{a^2+x^2} = \frac{1}{a} \arctan \frac{x}{a} + C$, right? $\endgroup$ – A.S Apr 1 '13 at 20:43
  • $\begingroup$ @AndrewSalmon Evidently not! $\endgroup$ – Adam Saltz Apr 1 '13 at 20:47
  • $\begingroup$ On thing at least is incorrect: Don't write $\ln\sec\theta+\tan\theta$ if you mean $\ln(\sec\theta+\tan\theta)$. Those are two different things. $\endgroup$ – Michael Hardy Apr 1 '13 at 21:32
  • $\begingroup$ @RonGordon hooray! $\sqrt2$ is rational :-) $\endgroup$ – obataku Apr 1 '13 at 22:40

Hint: you can cut your work considerably by using the trig substitution directly into the proper integral, and proceeding (no place for taking the square root of the denominator):

You have $$2\int_0^2 \frac{dx}{4+x^2}\quad\text{and NOT} \quad 2\int_0^2 \frac{dx}{\sqrt{4+x^2}}$$

But that's good, because this integral (on the left) is what you have and is already in in the form where it is appropriate to use the following substitution:

Let $x = 2 \tan \theta$, which you'll see is standard for integrals of this form.

As suggested by Andrew in the comments, we can arrive at his suggested result, and as shown in Wikipedia:

Given any integral in the form


we can substitute

$$x=a\tan(\theta),\quad dx=a\sec^2(\theta)\,d\theta, \quad \theta=\arctan\left(\tfrac{x}{a}\right)$$

Substituting gives us:

$$ \begin{align} \int\frac{dx}{{a^2+x^2}} & = \int\frac{a\sec^2(\theta)\,d\theta}{{a^2+a^2\tan^2(\theta)}} \\ \\ & = \int\frac{a\sec^2(\theta)\,d\theta}{{a^2(1+\tan^2(\theta))}} \\ \\ & = \int \frac{a\sec^2(\theta)\,d\theta}{{a^2\sec^2(\theta)}} \\ \\ & = \int \frac{d\theta}{a} \\ &= \tfrac{\theta}{a}+C \\ \\ & = \tfrac{1}{a} \arctan \left(\tfrac{x}{a}\right)+C \\ \\ \end{align} $$

Note, you would have gotten precisely the correct result had you not taken the square root of $\sec^2\theta$ in the denominator, i.e., if you had not evaluated the integral of the square root of your function.

  • $\begingroup$ But the original poster did use the substitution $x=2\tan\theta$. $\endgroup$ – Michael Hardy Apr 1 '13 at 21:34
  • $\begingroup$ Thank you all. I think I see my problem. The textbook says that these substitutions are for forms of $\sqrt{a^2 + x^2}$, then proceeds to give many examples of $\int\frac{dx}{\sqrt{a^2 + x^2}}$. Because this didn't have the radical, I was making the assumption that $x^2 = 2tan(\theta) => x = \sqrt{2tan(\theta)}$. I saw that what was given was the form I needed, but was lacking the one component that I thought was necessary. Learning can be so humbling. Thanks again. $\endgroup$ – Andrew Falanga Apr 2 '13 at 16:22
  • $\begingroup$ Andrew: We are all humbled when learning something new! (See my profile: first quote!) You did the hard part of your problem, and applied the substitution correctly: it's just that trying to make it conform to $\sqrt{\cdot}$ wasn't needed here. $\endgroup$ – amWhy Apr 2 '13 at 16:27

I think you correctly computed $$\sqrt{2}\int_0^2 \frac{dx}{\sqrt{4+x^2}}$$ but this has nothing to do with $2\int_0^2 \frac{dx}{4+x^2}$. They certainly aren't guaranteed to be equal. For example, $$\int_0^1 x^2 \; dx = 1/3$$ while $$ \int_0^1 x \;dx = 1/2.$$

Let's get back to your original question. You can use a trig substitution for this problem, but Andrew Salmon has proposed an easier way. $$2\int_0^2 \frac{dx}{4+x^2} = \frac{2}{4}\int_0^2 \frac{dx}{1 + (\frac{x}{2})^2}.$$ Now substitute $u = x/2$ and use the formula $\int \frac{dx}{1+x^2} = \arctan(x) + C$.

To use a trig substitution, note that $4+x^2 = (\sqrt{4+x^2})^2$, so $4+x^2$ is of the appropriate form. (Alternatively, be bold! and do the substitution without worrying about this.) Just use $x = 2\tan(\theta)$.


Consider the triangle with angle $\theta$, adjacent$=2$, opposite$=x$, and hypotenuse$=\sqrt{4+x^2}$. Then


Also, we need the $d\theta$ element, so let's take the simplest relation ${x\over 2}=\tan\theta$. Then $dx=2\sec^2\theta$. Thus,

$$\int_{x=-2}^2{4\over4+x^2}dx=\int\cos^2\theta\sec^2\theta d\theta=\theta=\tan^{-1}{x\over2}\bigg|_{x=-2}^2=\tan^{-1}(1)-\tan^{-1}(-1)={\pi\over2}.$$

Simply divide by 4.

  • $\begingroup$ oops, i forgot the 2 from dx=2sec theta $\endgroup$ – Brady Trainor Apr 2 '13 at 4:08

Let $x=2\tan\theta$ and $dx=2 \sec^2\theta$

$2\tan\theta = -2, 2$

$\theta = \frac{-\pi}{4}, \frac{\pi}{4}$



$\int_{\frac{-\pi}{4}}^\frac{\pi}{4}\frac{1}{2} d\theta$

$\frac{\pi}{8}-\frac{-\pi}{8} = \frac{\pi}{4}$

Sorry about the formatting. I'm just learning latex.

  • $\begingroup$ That's ok, I have to admit, the LaTeX formatting threw me for a loop too. In fact, I'd typed this up the night before I posted but couldn't get the parser to understand what I meant. I gave up in frustration. The next day, I discovered, it was because I need to have the arguments to the trig functions wrapped in parenthesis. lol (you'd think as a programmer, I'd have known that) $\endgroup$ – Andrew Falanga Apr 2 '13 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.