0
$\begingroup$

The table of integrals says that

\begin{equation*} \int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\arctan\frac{x}{a}+C \end{equation*}

where $C$ is a constant. What's wrong with my proof?

$$ \begin{align*} y &= \arctan\frac{x}{a} \\\ a\tan y&=x \\\ \tan y &= \frac{x}{a} \\\ \frac{d}{dx} \tan y &= \frac{d}{dx} \frac{x}{a} \\\ \sec^{2}y \frac{dy}{dx} &= \frac{1}{a} \\\ \frac{dy}{dx} &= \frac{1}{a} \frac{1}{\sec^{2}y} \\\ &= \frac{1}{a} \frac{1}{1 + \tan^{2} y} \\\ \frac{dy}{dx}&= \frac{1}{a} \frac{1}{1 + (\frac{x}{a})^{2}} \\\ \int \frac{dy}{dx}\\,dx &= \int \frac{1}{a} \frac{1}{1 + (\frac{x}{a})^{2}}\\,dx\\\ \arctan\frac{x}{a} &= \frac{1}{a} \int \frac{1}{1 + (\frac{x}{a})^{2}}\\,dx \\\ \end{align*} $$

Is it right so far? Any help appreciated!

  • 1
    If you are trying to show that the given function is indeed an antiderivative for $\frac{1}{a^2+x^2}$, then just take the derivative and check! If this is not your objective, then what you are you trying do?2010-11-07

2 Answers 2

3

Other than having unnecessary steps (why did you move $a$ to the left hand side in the second step, only to bring it back to the right hand side in the next step?), and missing the constant of integration on the left hand side in the last equation (as well as the missing "$dx$" which I fixed), nothing wrong so far. Now perform the operation inside the integral to get $\frac{a^2}{a^2+x^2}$, pull out the constant, simplify, and move it the left hand side to get the formula.

Of course, if what you want is to show that the integral of $\frac{1}{a^2+x^2}$ is really equal to $\frac{1}{a}\arctan(\frac{x}{a}) + C$, then why not simply differentiate the latter and see if you get the integrand?

  • 0
    Ohhk, so I can't perform the operation inside the integral yet (not yet on partial fractions (which is what that is, right?)) but I see what I have to do. The reason I have unnecessary steps in because I copied it straight from my paper to LaTeX. Thanks Arturo!2010-11-07
  • 0
    @G.P. Burdell: No, it's not a partial fraction; it's just algebra. The denominator is $1 + \frac{x^2}{a^2}$; rewrite $1$ as $\frac{a^2}{a^2}$, and write the denominator as a single fraction. Then you have $\frac{1}{(\frac{a^2+x^2}{a^2})}$, and writing this as a simple fraction gives you $\frac{a^2}{a^2+x^2}$. No "partial fractions" at all.2010-11-07
  • 2
    @G.P. Burdell: Telling me that your post has unnecessary steps because your original has unnecessary steps does not tell me *why* you have unnecessary steps. "It's elephants all the way down" doesn't really work as an answer.2010-11-07
1

A better way to prove it would be to let $\displaystyle x = a\tan{\theta}$, then $\displaystyle \frac{dx}{d\theta} = a\sec^2{\theta} \Rightarrow dx = a\sec^2{\theta}\;{d\theta}$. So:

$\displaystyle \int\frac{1}{x^2+a^2}\;{dx} = \int\frac{a\sec^2{\theta}}{a^2\tan^2{\theta}+a^2}\;{d\theta} = \frac{1}{a}\int\frac{a\sec^2{\theta}}{a(\tan^2{\theta}+1)}\;{d\theta}$

$\displaystyle = \frac{1}{a}\int\frac{a\sec^2{\theta}}{a\sec^2{\theta}}\;{d\theta} = \frac{1}{a}\int\;{d\theta} = \frac{1}{a}\theta+k = \frac{1}{a}\arctan\frac{x}{a}+k$.

  • 0
    @De Moivre: there are plenty of ways of *deriving* the formula. But to *prove* the formula, the most straightforward (and often best) method is simply to differentiate the answer. That is, in my opinion, the case here.2010-11-07
  • 0
    I don't understand, though: if differentiating the RHS of the formula is acceptable as a proof, why isn't interegrating the LHS so?2010-11-07
  • 0
    @De Moire: I did not say it was "not acceptable." What makes you think I did? I said that it usually the *most straightforward* and *best* way to do it the *easy* way: by differentiation. Differentiation only requires the application of formulas, it requires no inventiveness, changes of variables, clever ideas or tricks, etc.2010-11-07
  • 0
    I'm very sorry that I've misunderstood what you said. You're right of course about differentiation. It's indeed the easiest and the most straightforward way to do this.2010-11-07