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I'm currently learning integration techniques, and I often run into a problem; when i'm solving an integral, sometimes I need to use the natural log simplification and other times I need to use one of the inverse trig formulas, namely arctangent. I've done plenty of reading before I came here on when to use ln vs arctan, and perhaps my wording isn't correct(if so I apologize for that), but I haven't found any answers. Hopefully some of you fine folks can clarify things for me as my professor won't answer my emails.

For the following problem, i'm using the integration by parts technique:

$$\int{\arctan x} {\;dx} = x\;\arctan\;x \;- \int{\frac{x}{x^2\;+1}}\;dx$$

where u = $\arctan x$$\;\;du = \frac{1}{x^2+1}$ $\;\;dv = 1$ $\;\;v = x$

Then I use another substitution where $\;w = {x^2+1}$ and $\;\frac{dw}{2} = x\;dx$ for the integral left, giving me:

$$x\;\arctan\;x \:- \frac{1}{2}\int\frac{dw}{w}\;$$

After subbing back in, it looks to me like it's in the arctan format of $\frac{1}{a}\arctan\frac{x}{a}$ with $\;x^2+1$ in the denominator, but the answer is in fact:

$$x\;\arctan\;x \;-\; \frac{1}{2}(\ln|x^2+1|)+C$$

This sort of thing burns me often as i've done similar problems in the past that are the other way around. Is it because I used w substitution? Or perhaps i'm just not seeing something critical?

Thanks in advance for any help, and I apologize if my formatting sucks, i'm new to the site and I'm still learning.

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  • $\begingroup$ Did you intend to write $dx$ instead of $x$ in the numerator on the right hand side of your first formula? $\endgroup$ – Carlos Esparza Jun 8 '16 at 19:23
  • $\begingroup$ shoot i forgot the dx, my apologies. ill edit now $\endgroup$ – FuegoJohnson Jun 8 '16 at 19:24
  • $\begingroup$ I don't see anything wrong here...? You've made the substitution $w = x^2 + 1$, what do you do after that? The integral $$\int \frac{dw}{w}$$ is $\ln \lvert w \rvert$, which is exactly what you're looking for. Have you subbed back in $w = x^2 + 1$ before evaluating the integral? In that case you have $$\int \frac{1}{x^2 + 1}\ d(x^2 + 1)$$. $\endgroup$ – Edward Evans Jun 8 '16 at 19:34
  • $\begingroup$ True, but when I first looked at it, I thought I needed to use the arctan setup, and I was of course wrong. I've done similar problems in the past though that I thought was a natural log but was in fact arctan. This is the issue I keep running into, when its appropriate to use what... $\endgroup$ – FuegoJohnson Jun 8 '16 at 19:36
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    $\begingroup$ Thank you so much, you clarified exactly what I needed to know. Basically, by subbing using w, it eliminated the arctan possibility which is what I was confused about. So when it's straight up as you posted above, that's when ill use arctan. Thanks again! $\endgroup$ – FuegoJohnson Jun 8 '16 at 19:43
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Based off Ed_4434's excellent input(and since he chose to answer my question in the comments section) I figured i'd post a concrete answer for anyone else who may have a similar question to mine:

In the case of my problem above, since I used a w substitution, it eliminates the possibility of $\arctan\;x + C$ as an answer. Even though it looks similar to $\arctan$ in the form of: $$\int\frac{1}{w}$$ where $w = x^2+1$, it is in fact a natural log due to the substitution method I used. If the integral simplified to: $$\int\frac{1}{x^2+1}dx$$ without the use of substitution(and with respect to x), then the answer would in fact be $\arctan\;x + C$.

In short, if an integral simplifies to something that looks like the form of $\frac{1}{a}\arctan\frac{x}{a}$(without substitution methods) then use $\arctan$ . Otherwise, if substitution is used, the answer would be a natural log situation(like the correct answer I posted in my original question). Hope this helps someone.

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