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u-Substitution

Summary: u-Substitution
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I. u-Substitution in Indefinite Integrals

If you were asked to evaluate the integral $\int \! e^{\boxdot} \, d\boxdot,$

you would probably guess that the answer is
$$\int \! e^{\boxdot} \, d\boxdot = e^{\boxdot} + C$$
no matter what exactly is in the $\boxdot$ box. You would be correct!




Similarly, these integrals are all correct:
\begin{align*}
\int \! \cos \left(\boxdot \right) \, d\boxdot &= \sin \left(\boxdot \right) + C \\[8px]
\int \! \sin \left(\boxdot \right) \, d\boxdot &= -\cos \left(\boxdot \right) + C \\[8px]
\int \! \left(\boxdot \right)^2 \, d\boxdot &= \dfrac{1}{3}\left(\boxdot \right)^3 + C
\end{align*}
What’s crucial in each instance above is that the $”\boxdot”$ is identical in both the function you’re integrating ($e^{\boxdot}$, or $\cos \left(\boxdot \right)$, …) and the $d\boxdot.$

For instance, this is correct:
$$\int \! e^{ \bbox[yellow,5px]{5x}} \, d( \bbox[yellow,5px]{5x}) = e^{ \bbox[yellow,5px]{5x}} + C$$
By constrast, if we have $d(x)$ instead of $d(5x),$ then we don’t immediately know the integral:
$$ \int \! e^{ \bbox[yellow,5px]{5x}} \, d\bbox[orange,5px]{x} \ne e^{ \bbox[yellow,5px]{5x}} + C$$
We can check that the result on the right-hand side of the equation isn’t correct because if we take its derivative, the Chain rule gives us an extra factor of 5, and so we we don’t get back the integrand:
$$\dfrac{d}{dx}\left(e^{5x} + C \right) = e^{5x} \cdot 5 + 0 \ne e^{5x} $$
We can, however, turn the integral $\int \! \,e^{5x} dx$ into one we can evaluate easily by making what’s known as a u-substitution. The process essentially consists of guessing—yes, guessing—what would be a useful variable to integrate with respect to, and then convert the integral you have into one that’s entirely in terms of that variable.



That probably sounds abstract, we know. As usual, it’s easiest (and best) to show you how it works by working through a few examples, and then you can work through many problems to try it out for yourself. We promise that with just a little practice, you’ll get good at turning the integrals you’re given into ones you already know how to evaluate.



The following example illustrates.


Example 1

Find $\int \! e^{5x} \, dx.$



Solution.

You might think, “I know that $\int \! e^u \, du = e^u + C,$” so let’s try $u = 5x$ and see what happens.
$$u = 5x$$
Then
\begin{align*}
du &= 5 \, dx \\[8px]
\implies dx &= \dfrac{1}{5}du
\end{align*}
Now let’s make those substitutions into the integral:
\begin{align*}
\int \! e^{(\overbrace{5x}^u)} \, \overbrace{dx}^{\frac{1}{5}du} &= \int \! e^u \, \left(\dfrac{1}{5} du \right) \\[8px]
&= \dfrac{1}{5} \int \! e^{\bbox[yellow,5px]{u}} \, d\bbox[yellow,5px]{u} \\[8px]
\end{align*}
$$\text{Ah, magic: we know that integral!}$$
\begin{align*}
\phantom{\int \! e^{(\overbrace{5x}^u)} \, \overbrace{dx}^{\frac{1}{5}du} } &= \dfrac{1}{5}e^{\bbox[yellow,5px]{u}} + C \\[8px]
&= \dfrac{1}{5}e^{5x} + C \quad \cmark
\end{align*}
Notice that in the last line we merely substituted back $5x = u:$ since the original problem was in terms of $x,$ our final answer also needs to be in terms of $x$ instead of $u.$



Let’s check that our answer is correct:
\begin{align*}
\dfrac{d}{dx}\left(\dfrac{1}{5}e^{5x} + C \right) &= \dfrac{1}{5}\dfrac{d}{dx}\left(e^{5x} \right) + 0 \\[8px]
&= \dfrac{1}{5}\left(e^{5x}\cdot 5 \right) \\[8px]
&= e^{5x} \quad \cmark
\end{align*}
FAQ

Why doesn’t $\dfrac{1}{5}\int \! e^u \, du = \dfrac{1}{5}\left(e^u + C \right)$? That is, why is your answer “$+C$” instead of “$+\dfrac{1}{5}C$”?

Answer: The constant $+C$ is a placeholder for some constant, and we don’t know or care what it is. By convention, we thus always write $+C$ rather than $+\dfrac{1}{5}C,$ or $-C,$ or $2C,$ or anything else.

End Example 1.


Let’s consider another example.

 


Example 2.

Find $\int \! \cos(x^2) \,x \, dx.$


Solution.

You might think, “I know that $\int \! \cos(u) \, du = \sin (u) + C,$ so let’s try $u = x^2$ and see what happens.”

If you were thinking something similar, you’re on the right track!



That is, let
$$u = x^2$$
Then
\begin{align*}
du &= 2x \, dx \\[8px]
\implies x \, dx &= \dfrac{1}{2}du
\end{align*}
Note that it’s fortunate that the original integrand has that “extra” $x$ in it: we need that $x$ in order to make the substitution $x \, dx = \dfrac{1}{2}du.$ In fact if that $x$ weren’t there, we’d be stuck and couldn’t proceed. (But then you wouldn’t be given the integral $\int \! \cos(x^2) \, dx,$ because we can’t solve it with the tools we have. We need that “extra” $x$ there.)



Let’s make the substitution $u = x^2$ into our original integral and see what happens:
\begin{align*}
\int \! \cos(\overbrace{x^2}^u) \,(\overbrace{x \, dx}^{\frac{1}{2}du}) & = \int \! \cos (u) \left( \dfrac{1}{2} du\right) \\[8px]
&= \dfrac{1}{2} \int \! \cos(\bbox[yellow,5px]{u}) \, d\bbox[yellow,5px]{u} \\[8px]
\end{align*}
$$\text{Ah, again we know that integral: }$$
\begin{align*}
\phantom{ \int \! \cos(\overbrace{x^2}^u) \,(\overbrace{x \, dx}^{\frac{1}{2}du})} &= \dfrac{1}{2}\sin(\bbox[yellow,5px]{u}) + C \\[8px]
&= \dfrac{1}{2} \sin \left(x^2 \right) + C \quad \cmark
\end{align*}
Again in the last step we substituted for $u$ in terms of $x\: \left(u = x^2 \right)$ so our final answer is in terms of $x$ instead of $u.$



Let’s check that our answer is correct:
\begin{align*}
\dfrac{d}{dx}\left(\dfrac{1}{2} \sin \left(x^2 \right) + C \right) &= \dfrac{1}{2} \dfrac{d}{dx}\sin(x^2) + 0 \\[8px]
&= \dfrac{1}{2}\cos(x^2)\cdot \dfrac{d}{dx}\left(x^2 \right) \\[8px]
&= \dfrac{1}{2}\cos(x^2)\cdot (2x) \\[8px]
&= \cos(x^2) \: x \quad \cmark
\end{align*}
End Example 2.


The upshot: As you can see, making a u-substitution can quickly turn an integral you don’t immediately know into one that you do. To do so, guess what a good choice for $u$ is, and then see what happens.


 

II. u-Substitution in Definite Integrals

If you’re given a definite integral (with limits of integration), then it’s easiest to convert those $x$ values into their equivalent $u$ values and then complete the calculation in terms of $u$. The following example illustrates.

Example 3.

Find $\int_0^1 \! e^{5x} \, dx.$




Solution.

As in Example 1 above, let $u = 5x:$
\begin{align*}
u &= 5x \\[8px]
du &= 5 \, dx \\[8px]
\implies dx &= \dfrac{1}{5}du
\end{align*}
We must also convert the limits of integration to be in terms of $u:$
\begin{align*}
u &= 5x \\[8px]
\text{When }x=0: \quad u &= 5(0) = 0 \\[8px]
\text{When }x=1: \quad u &= 5(1) = 5
\end{align*}
Now let’s make those substitutions into the integral, changing the limits of integration too:
\begin{align*}
\int_0^1 \! e^{(\overbrace{5x}^u)} \, \overbrace{dx}^{\frac{1}{5}du} &= \int_\bbox[yellow,5px]{0}^\bbox[yellow,5px]{5} \! e^u \, \left(\dfrac{1}{5} du \right) \\[8px]
&= \dfrac{1}{5} \int_0^5 \! e^{u} \, du \\[8px]
&= \dfrac{1}{5}\left[ e^{u}\right]_0^5 \\[8px]
&= \dfrac{1}{5}\left[e^5 – e^0 \right] \\[8px]
&= \dfrac{1}{5}\left[e^5 – 1 \right] \quad \cmark
\end{align*}
End Example 3.


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There are many more example problems below so you can get the hang of how to do u-substititions.

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