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I'm trying to solve

$$\int_{-\infty}^{\infty}{\frac{1}{(4+x^2)\sqrt{4+x^2}} \space dx}$$

By substituting $x=2\tan{t}$. I get as far as:

$$\int_{x \space = -\infty}^{x \space = \infty}{\frac{1}{(4+(\underbrace{2\tan{t}}_{x})^2)\sqrt{4+(\underbrace{2\tan{t}}_{x})^2}} \cdot \underbrace{2(1+\tan^2t) \space dt}_{dx}} = \dots$$

$$\dots = \frac{1}{4} \cdot \int_{t \space = -\infty}^{t \space = \infty}{\frac{1}{\sqrt{1+\tan^2t}} \space dt}$$

Now what? Have I done anything wrong? I don't see how I could continue from now on.

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    $\begingroup$ $1+\tan^2x=\sec^2x$ (follows from $\sin^2x+\cos^2x=1$) $\endgroup$
    – yoyo
    Commented Dec 20, 2011 at 20:32
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    $\begingroup$ Your steps are correct, except for a small thing: when you do a substitution (here, for $x$ in terms of $t$), you should remember to calculate the limits for $t$ and plug that in the new integral. In this example, the limits for $t$ are $-\frac{\pi}{2}$ and $\frac{\pi}{2}$; do you see why? // To proceed, one could use the identity: $1 + \tan^2 x = \sec^2 x = \frac{1}{\cos^2 x}$. $\endgroup$
    – Srivatsan
    Commented Dec 20, 2011 at 20:33
  • $\begingroup$ I'm not good at limits so I left those values the same until I'm done with the substitution. We haven't learned anything called $sec(x)$ at uni, I should probably look it up. $\endgroup$
    – dimme
    Commented Dec 20, 2011 at 20:36
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    $\begingroup$ Do look it up! Without it you could note that $1+\tan^2 x=1+\frac{\sin^2 x}{\cos^2 x}=\frac{\sin^2 x+\cos^2 x}{\cos^2 x}=\frac{1}{\cos^2 x}$. $\endgroup$ Commented Dec 20, 2011 at 20:43
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    $\begingroup$ "Solve" is the wrong word. You don't solve an integral; you evaluate an integral. $\endgroup$ Commented Dec 20, 2011 at 21:41

1 Answer 1

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Your substitution is the way to go in this problem. To complete the substitution however, I recommend that you calculate the new limits for $t$. An alternate approach is to postpone this till the end, but this approach invariably ends up confusing if there's a string of substitutions. In this example, as $x \to -\infty$, $t = \arctan x \to -\frac{\pi}{2}$; similarly as $x \to \infty$, we have $t \to \frac{\pi}{2}$. Therefore, after substitution, the integral becomes $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{1+\tan^2 t}} \, dt. $$


To proceed further, you need to use the following cousin of the Pythagorean theorem: $$ 1 + \tan^2 t = \sec^2 t . $$ This identity is extremely important and useful in practice -- not the least in manipulating integrals like this. One should be reminded of this identity whenever one comes across an expression like $\sqrt{1 + \tan^2 t}$ or $\sqrt{1+ x^2}$. By the way, the proof of this identity is based on the standard Pythagorean theorem: $$ 1+ \tan^2 t = 1 + \frac{\sin^2 t}{\cos^2 t} = \frac{\cos^2 t + \sin^2 t}{\cos^2 t} = \frac{1}{\cos^2 t} = \sec^2 t. $$ From this, it follows that $$ \frac{1}{\sqrt{1 + \tan^2 t}} = |\cos t|. $$

Thus the integral becomes $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\cos t| \, dt. $$ Can you take it from here?

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