Struggling with an integral with trig substitution

Question

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

Answer 1

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

$$\int\frac{dx}{{a^2+x^2}}$$

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.

Answer 2

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 $x = 2\tan(\theta)$.

Answer 3

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

$${2\over\sqrt{4+x^2}}={\text{adj}\over\text{hyp}}=\cos\theta.$$

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.

Answer 4

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

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

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

$\int_{\frac{-\pi}{4}}^\frac{\pi}{4}\frac{2\sec^2\theta}{4+4\tan^2\theta}d\theta$

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

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

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

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