calculus - Struggling with an integral with trig substitution

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

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.

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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.