How to see a U-Sub

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We give some hints and tips on how to see and predict where a U-Sub might help.

One of the most difficult aspects of the $u$ -substitution is knowing what to substitute. This can be exacerbated by the fact that instructors, by virtue of having taught $u$ -subs for years and practiced them hundreds or even thousands of times, always seem to know the exactly perfect substitution to make right away. But for a student learning this technique for the first time, this is a fairly unreasonable goal. In this segment we will discuss how someone new to this technique might figure out how solve complicated integrals that require elaborate $u$ -substitutions.

Video Lecture

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The best thing to remember is that you don’t need to do everything in one substitution. Instructors often perform one massive substitution because they have the experience to see the exact (often crazy/elaborate) substitution that perfectly converts the integrand into something easy to compute, and it saves time when presenting problems in a classroom. But there is absolutely no mathematical reason you need to do everything in a single substitution. Consider the following example:

$\int 2x\sqrt {e^{x^2 + 1} - 2}e^{x^2+1}\dx$

The best way to tackle a complicated expression like this, is one piece at a time. Look for something that looks like it might be a viable substitution and go with that. Don’t worry about catching everything, or about trying to make sure the result after the substitution is something you can integrate right away. Do a substitution, then revisit the result as if it is a new integral and see if there is an antiderivative, or another substitution that makes sense.

Let’s go through our example from start to finish. First we might notice that a part of the exponent $x^2$ might make a good substitution because there is a $2x$ multiplier on the outside. So let’s start with that - let $u = x^2$ , then $\du = 2x\dx$ , and so $\dx = \frac {\du }{2x}$ . Now we can substitute into our original integral to get:

$\displaystyle \int 2x\sqrt {e^{x^2 + 1} - 2}e^{x^2+1}\dx$	$=\displaystyle \int 2x\sqrt {e^{u+1} - 2}e^{u+1}\frac {\du }{2x}$
	$=\displaystyle \int \sqrt {e^{u+1} - 2}e^{u+1}\du$

But, now that we look at the new exponent $u+1$ , that $+1$ isn’t really doing anything other than making things difficult. So let’s do a substitution to get rid of that. Thus we can do another substitution; $v = u+1$ . Then $\dv = \du$ and we now have:

$\displaystyle \int 2x\sqrt {e^{x^2 + 1} - 2}e^{x^2+1}\dx$	$=\displaystyle \int 2x\sqrt {e^{u+1} - 2}e^{u+1}\frac {\du }{2x}$
	$=\displaystyle \int \sqrt {e^{u+1} - 2}e^{u+1}\du$
	$=\displaystyle \int \sqrt {e^{v} - 2}e^{v}\dv$

Ok, now the exponent on $e$ is better. But looking at it further, that $e^v$ is inside the square root, and also multiplying the outside. We could probably do a substitution for that, so let’s try $w=e^v$ and see what happens. So, $\dw = e^v \dv$ , and $\dv = \frac {\dw }{e^v}$ . This gives us;

$\displaystyle \int 2x\sqrt {e^{x^2 + 1} - 2}e^{x^2+1}\dx$	$=\displaystyle \int 2x\sqrt {e^{u+1} - 2}e^{u+1}\frac {\du }{2x}$
	$=\displaystyle \int \sqrt {e^{u+1} - 2}e^{u+1}\du$
	$=\displaystyle \int \sqrt {e^{v} - 2}e^{v}\dv$
	$=\displaystyle \int \sqrt {w - 2}e^{v}\frac {\dw }{e^v}$
	$=\displaystyle \int \sqrt {w - 2}\dw$

Again we notice here that the $-2$ part of $w-2$ doesn’t really do much here, so we can try another substitution to get rid of it. Let’s try $s = w-2$ . Then $\ds = \dw$ and we have;

$\displaystyle \int 2x\sqrt {e^{x^2 + 1} - 2}e^{x^2+1}\dx$	$=\displaystyle \int 2x\sqrt {e^{u+1} - 2}e^{u+1}\frac {\du }{2x}$
	$=\displaystyle \int \sqrt {e^{u+1} - 2}e^{u+1}\du$
	$=\displaystyle \int \sqrt {e^{v} - 2}e^{v}\dv$
	$=\displaystyle \int \sqrt {w - 2}e^{v}\frac {\dw }{e^v}$
	$=\displaystyle \int \sqrt {w - 2}\dw$
	$=\displaystyle \int \sqrt {s}\ds$

Finally! We can do this integral! Specifically $\displaystyle \int \sqrt {s}ds = \frac {2}{3}s^{\frac {3}{2}}+C$ . Now we can unravel all our substitutions:

$\displaystyle \frac {2}{3}s^{\frac {3}{2}}+C$	$=\displaystyle \frac {2}{3}(w-2)^{\frac {3}{2}}+C$	$s = w-2$ .
	$=\displaystyle \frac {2}{3}\left (\left (e^v\right )-2\right )^{\frac {3}{2}}+C$	$w = e^v$ .
	$=\displaystyle \frac {2}{3}\left (\left (e^{u+1}\right )-2\right )^{\frac {3}{2}}+C$	$v=u+1$ .
	$=\displaystyle \frac {2}{3}\left (\left (e^{x^2+1}\right )-2\right )^{\frac {3}{2}}+C$	$u=x^2$ .

So, our end result is $\displaystyle \int 2x\sqrt {e^{x^2 + 1} - 2}e^{x^2+1}\dx = \frac {2}{3}\left (\left (e^{x^2+1}\right )-2\right )^{\frac {3}{2}}+C$ .

Now, it’s true we could have done the entire thing in one step if we happened to see the substitution $u = e^{x^2+1}-2$ as our first substitution - and if you managed to see that, awesome! But most students, especially those first learning and using $u$ -substitutions, would not have seen that. Importantly, a student may have seen some better substitutions along the way in the above example (we purposefully did a lot of substitutions for demonstration), but the point is that although doing more complicated substitutions might make the work faster, it won’t make the answer any more correct!

Until you get the experience and confidence to see and try more complex $u$ -substitutions, rest assured that it is not only acceptable, but a good idea, to do multiple substitutions to handle more complicated cases, rather than spending a lot of time trying to figure out the “perfect” substitution. Moreover, attempting to do a large or complicated substitution often leads to more computational errors, whereas multiple simpler substitutions usually results in fewer computational errors.

1 : When doing substitutions, especially when first learning the $u$ -substitution method for integration, it is best to...

Do as few substitutions as possible to reduce the number of steps and compute stuff faster. Go out of your way to do as many substitutions as possible to make them smaller. Do the substitutions you see, even if they are small, and use as many (or as few) as you are comfortable with and need. Don’t go out of your way trying to use more or fewer substitutions just because you think you are suppose to. Look up the best substitution to use online.

The big takeaway from this is that often those demonstrating $u$ -substitutions for students have years of experience and have done this kind of substitution hundreds or thousands of times. As a result, it’s natural for them to do complicated large substitutions all at once, because for them they aren’t especially difficult to see or do. But this gives the false impression to students that they are suppose to do substitutions all at once - searching for the “perfect” single substitution. This is not only untrue, but generally a really bad idea, leading to more computational errors and frustration for the student. Remember that it is not just ok, but actually a good idea, to do substitutions piecemeal - do whatever substitution you see that might make the problem simpler, one after another, until you can get to something you can integrate!