# Integration by parts, the cases when it does not matter what $u$ and $dv$ we choose.

I was reviewing Integration By Parts on Brilliant.org where an example they use is

$$\int x \ln x \;dx$$ Let $u=\ln x$ and $dv=x\;dx$ such that \begin{align} \int x \ln x\;dx&\;=\;\frac 12x^2\ln x\;-\;\frac 12\int \frac{x^2}x\;dx\\ \\ &\;=\; \frac 12x^2\ln x\;-\;\frac 14x^2\;+\;C \\ \end{align}

Personally, I would have solved it by letting $u=x$ and $dv=\ln x\;dx$ such that \begin{align} \int x \ln x\;dx&\;=\;x^2\left(\ln x -1\right)\;-\;\int x \ln x\;dx\;+\;\int x\;dx\\ \\ &\;=\; \frac 12\left[x^2\left(\ln x - 1\right)\;+\;\frac 12x^2\right]\;+\;C\\ \\ &\;=\; \frac 12x^2\ln x\;-\;\frac 14x^2\;+\;C \\ \end{align}

So both choices for $u$ and $dv$ yield a correct result that also is the same both ways up to a constant and I would like to know if that is trivial or what that means.

Is there something interesting about integrals such as $\int x\ln x\;dx$ where it does not matter what $u$ and $dv$ we choose to perform Integration By Parts and would there be other examples of this?

## Edit

This edit is to add to the main post the integral

$$\int \cos(ax)e^{bx}\;dx$$

from @Michael Hardy's comment

Letting $u=\cos(ax)$ and $dv=e^{bx}dx$ yields

\begin{align} \\ \int \cos(ax)e^{bx}\;dx&\;=\;\underbrace{\frac 1b\cos(ax) e^{bx}}_A\;+\;\underbrace{\frac ab}_B \int \sin(ax)e^{bx}\;dx\\ \\ &\;=\;A\;+\;B \left[\frac 1b\sin(ax)e^{bx}\;-\;B\int \cos(ax)e^{bx}\;dx\right]\\ \\ &\;=\;\frac 1{1+B^2}\left(A\;+\;\frac a{b^2}\sin(ax)e^{bx}\right)\;+\;C\\ \\ &\;=\;\frac {e^{bx}}{a^2+b^2}\left(b\cos(ax)+a\sin(ax)\right)\;+\;C\\ \\ \end{align}

which is the same result as when letting $u=e^{bx}$ and $dv=\cos(ax)\;dx$ such that

\begin{align} \\ \int \cos(ax)e^{bx}\;dx&\;=\;\underbrace{\frac 1a\sin(ax) e^{bx}}_A\;-\;\underbrace{\frac ba}_B \int \sin(ax)e^{bx}\;dx\\ \\ &\;=\;A\;-\;B \left[-\frac 1a\cos(ax)e^{bx}\;+\;B\int \cos(ax)e^{bx}\;dx\right]\\ \\ &\;=\;\frac 1{1+B^2}\left(A\;+\;\frac b{a^2}\cos(ax)e^{bx}\right)\;+\;C\\ \\ &\;=\;\frac {e^{bx}}{a^2+b^2}\left(a\sin(ax)\;+\;b\cos(ax)\right)\;+\;C\\ \\ \end{align}

## Edit (2)

This edit is to add the $3$ following integrals:

$$\int x \tan^{-1}x\;dx$$ $$\int x \cos^{-1}x\;dx$$ $$\int x \sin^{-1}x\;dx$$

that I got from @User8128's answer, after reading "logarithms and inverse trig functions become algebraic functions when differentiated".

• $\int x \tan^{-1}x\;dx$ letting $u=x$ and $dv=\tan^{-1}x\;dx$

\begin{align} \int x \tan^{-1}x\;dx&\;=\;\underbrace{x^2\tan^{-1}x-\frac x2\ln|1+x^2|}_\alpha-\int x\tan^{-1}x\;dx+\frac 12\int\ln|1+x^2|\;dx\\ \\ &\;=\;\frac 12\left[\alpha+\frac 12 x \ln|1+x^2|-2x+2\tan^{-1}x\right]+C\\ \\ &\;=\;\frac {x^2}2\tan^{-1}x-x+\tan^{-1}x+C\\ \\ \end{align}

• $\int x \tan^{-1}x\;dx$ letting $u=\tan^{-1}x$ and $du=x\;dx$

\begin{align} \int x \tan^{-1}x\;dx &\;=\;\frac {x^2}2 \tan^{-1}x-\frac 12 \int \frac {x^2}{1+x^2}\;dx\\ \\ &\;=\;\frac {x^2}2 \tan^{-1}x-x+\tan^{-1}x+C\\ \\ \end{align}

• $\int x \cos^{-1}x\;dx$ letting $u=x$ and $dv=\cos^{-1}x\;dx$

\begin{align} \int x \cos^{-1}x\;dx&\;=\;\underbrace{x^2\cos^{-1}-x\sqrt{1-x^2}}_\beta -\int x \cos^{-1}x\;dx+\int \sqrt{1-x^2}\;dx\\ \\ &\;=\;\frac 12 \left[\beta+\frac 12 \left(x\sqrt{1-x^2}+\sin^{-1}x\right)\right]+C\\ \\ &\;=\;\frac {x^2}2 \cos^{-1}x-\frac 14 x\sqrt{1-x^2}+\frac 14 \sin^{-1}x+C\\ \\ \end{align}

• $\int x \cos^{-1}x\;dx$ letting $u=\cos^{-1}x$ and $dv=x\;dx$

\begin{align} \int x\cos^{-1}x\;dx&\;=\;\frac {x^2}2\cos^{-1}x+\frac 12 \int \frac {x^2}{\sqrt{1-x^2}}\;dx\\ \\ &\;=\;\frac {x^2}2 \cos^{-1}x-\frac 14 x\sqrt{1-x^2}+\frac 14 \sin^{-1}x+C\\ \\ \end{align}

• $\int x \sin^{-1}x\;dx$ letting $u=x$ and $dv=\sin^{-1}x\;dx$

\begin{align} \int x \sin^{-1}x\;dx&\;=\;\underbrace{x^2\sin^{-1}+x\sqrt{1-x^2}}_\gamma -\int x \sin^{-1}x\;dx-\int \sqrt{1-x^2}\;dx\\ \\ &\;=\;\frac 12 \left[\gamma-\frac 12 \left(x\sqrt{1-x^2}-\sin^{-1}x\right)\right]+C\\ \\ &\;=\;\frac {x^2}2 \sin^{-1}x+\frac 14 x\sqrt{1-x^2}-\frac 14 \sin^{-1}x+C\\ \\ \end{align}

• $\int x \sin^{-1}x\;dx$ letting $u=\sin^{-1}x$ and $dv=x\;dx$

\begin{align} \int x\sin^{-1}x\;dx&\;=\;\frac {x^2}2\sin^{-1}x-\frac 12 \int \frac {x^2}{\sqrt{1-x^2}}\;dx\\ \\ &\;=\;\frac {x^2}2 \sin^{-1}x+\frac 14 x\sqrt{1-x^2}-\frac 14 \sin^{-1}x+C\\ \\ \end{align}

I see that in the case of $\cos^{-1}x$ and $\sin^{-1}x$, one choice of $u$ and $dv$ either turns the inverse trig function into an algebraic function right away (i.e. $d\left(\cos^{-1}x\right)=\frac{-1}{\sqrt{1-x^2}}\;dx$) by differentiation or the other choice of $u$ and $dv$ makes the original function "re-appear" along with an algebraic function (i.e. $\int \cos^{-1}x\;dx = x\cos^{-1}x-\sqrt{1-x^2}+C$) by integration.

In the case of $\tan^{-1}x$, a $\log$ appears if we integrate the function but then $\log$s turn into algebraic functions upon derivation as well or make the original function "re-appear" along with an algebraic function upon integration (i.e $\int \ln x\;dx = x\ln x - x +C$ ) and so we get the same result.

I understand that this is not special but I'm still trying to fully wrap my head around the reason why these integrals are such that when we apply Integration By Parts, any choice for $u$ and $dv$ yield correct results that are also identical to each other.

## Edit (last)

I got "Calculus: An Intuitive and Physical Approach" by Morris Kline and at the end of chapter $14$ on Further Techniques of Integration in section $6$ on The Use of Tables he says:

"[...] One answer would be to study more techniques or seek to discover a new one. However, the number of techniques and special tricks is quite extensive. It is neither wise nor efficient to spend months or years on what is really an incidental process or means to an end at the expense of the acquisition of more significant knowledge."

So, I will follow this advice and (for now) be content with the fact that I understand much better how integrals are obtained, avoid wasting time on knowing why some peculiar results of this "incidental process or means to an end" do occur and be going on my merry mathematical way.

• One other case in which it doesn't matter is things like $$\int \cos(ax) e^{bx} \,dx.$$ Integrating by parts TWICE yields $$A + B\int\cos(ax)e^{bx}\,dx$$ where $A$ is a function of $x$ and $B$ is a constant. Then you get $$\int \cos(ax)e^{bx}\,dx - B\int\cos(ax)e^{bx}\,dx = A + \text{constant}$$ and then divide both sides by $1-B. \qquad$ – Michael Hardy Jun 19 '17 at 15:31
• Integration by parts is always symmetric. I think that there is nothing special about this. What really matters if you can actually compute the integral in either way. – Siminore Jun 23 '17 at 14:57
• In both examples provided, at least one direction of integration by parts leads to the reappearance of the original integral. I think this makes them distinct from others where the concept fails. – Dunham Jun 23 '17 at 15:10

There can't really be any general rule for choosing $u$ and $v$ for integration by parts but there are hierarchies that make sense most of the time. The rule I like to use for choosing $u$ (this is not my rule; it is well documented if you google it) is LIATE (logarithm - inverse trig - algebraic - trig - exponential). That is, if there are logarithms present, you should first try to eliminate them with $u$; next you should look for inverse trig functions, then algebraic functions (polynomials, rational functions) and so forth. My general philosophy is choose $u$ which becomes "better" when you differentiate it. Here by better, I mean something like "more elementary." For example, logarithms and inverse trig functions become algebraic functions when differentiated; polynomials become lower order polynomials when differentiated. This makes these good candidates for $u$. For $dv$, I try to choose something which "doesn't get much worse" when anti-differentiated. This makes trig functions and exponentials good candidates for $dv$ since, upon anti-differentiation they remain trig functions and exponentials (respectively). This sort of thinking explains why we should choose $u = x$ and $dv = \cos(x) dx$ when integrating $$\int x \cos(x) dx.$$ It also explains why the choice is arbitrary for integral like $$\int \cos(ax) e^{bx} dx.$$ Neither of these get better upon differentiation and neither get worse upon anti-differentiation so you just choose one to be $u$ and one to be $dv$, perform IBP twice and arrive at the result; no matter which way you've chosen, you don't do any more or less work.

• After reading "logarithms and inverse trig functions become algebraic functions when differentiated" I tried $\int x \tan^{-1}x \;dx$ and it also yield identical results both ways... Anyways, thanks for the insight! – user408202 Jun 23 '17 at 19:31

Lets say you are trying to integrate $f(x) = u\frac{dv}{dx} = \mu\frac{d\nu}{dx}$. Then

\begin{align} \mu\nu -\int\nu\,d\mu = \int f(x)\,dx = uv-\int v\,du \end{align}

or, to put it more simply, you are evaluating the same integral by two different methods, of course they will always both give you the same result (up to a constant), otherwise your original integral would not have been well defined.

• I don't see what you mean by "two different methods" I am only applying Integration By Parts and it's the choice of $u$ and $dv$ that is not important here; as opposed to $\int x \cos(x)\;dx$ for example, which is well defined, yet choosing $u=\cos(x)$ and $dv=x\;dx$ and integrating by parts repeatedly does not work while obviously letting $u=x$ and $du=\cos(x)\;dx$ yields $$\int x\cos(x)\;dx\;=\;x\sin(x)+\cos(x)+C\;.$$ – user408202 Jun 20 '17 at 16:44
• @TheQuestion: I think the miscommunication here is that you don't really explain what you mean by "work". The point of this answer is to emphasize that every choice of splitting the integrand into $u$ and $dv$ works, in the sense that it gives a result that is true and correct. This interpretation of the meaning of "work" is further justified by your remark on getting the same result both ways. – user14972 Jun 23 '17 at 15:10
• @TheQuestion: I get the impression that what you mean by "work" is something like the immediate result of IBT being something useful -- but you never really say anything like that; I only infer that by reading between the lines. – user14972 Jun 23 '17 at 15:12
• @Hurkyl Now that you point it out I see that my choice of word is not clear, thank you. I really mean "work" in the sense of getting a true and correct result that also happens to be the same result both ways. – user408202 Jun 23 '17 at 15:25