Why u-Substitution is So Awesome

Students are introduced to “ $u$ -substitution” as yet another exam-question solving technique. That is not wrong, and does make for many challenging questions.

But the reason it is so is massive—the chain rule helps us discover new integrals that we would not have known otherwise.

Let’s first state $u$ -substitution, or what I’d like to call, the reverse chain rule.

Theorem 1 (Reverse Chain Rule). Let $f,g$ be (known) differentiable functions. Then letting $u := g(x),$

$\displaystyle \int f'(g(x))g'(x)\, \mathrm dx = f(g(x))+C = f(u) + C = \int f'(u)\, \mathrm du.$

Proof. By the chain rule,

$\displaystyle \frac{\mathrm d}{\mathrm d x} (f(g(x))) = f'(g(x)) g'(x) \quad \Rightarrow \quad \int f'(g(x))g'(x)\, \mathrm dx = f(g(x)) + C.$

The power of the reverse chain rule arises not from its theory—it’s right there—but from its applications. For instance, whenever we have an integrand (i.e. the function being integrated) of the form $f'/f$ , we obtain a very useful integral.

Theorem 2. Let $f$ be a known differentiable function. Then

$\displaystyle \int \frac{f'(x)}{f(x)}\, \mathrm dx = \ln|{f(x)}| + C.$

Proof. While the result is straightforward, a computation helps us evaluate the integral more conveniently. Let $u = f(x)$ . Then

$\displaystyle \frac{\mathrm du}{\mathrm dx} = f'(x) \quad \iff \quad \mathrm du = f'(x)\, \mathrm dx .$

Here, the notation on the right-hand side is purely formal and not a well-defined mathematical notion. However, this helps us compute the $u$ -substitution:

$\begin{aligned} \displaystyle \int \frac{f'(x)}{f(x)}\, \mathrm dx &= \int \frac{1}{f(x)}\, f'(x)\, \mathrm dx \\ &= \int \frac 1u\, \mathrm du \\ &= \ln|u| + C \\ &= \ln|{f(x)}| + C. \end{aligned}$

This helps us integrate $\tan$ . Recall that $\tan' = \sec^2$ . It is definitely not the case that

$\displaystyle \int \tan(x)\, \mathrm dx = \sec^2(x) + C.$

But, we do have the following result:

Corollary 1. $\displaystyle \int \tan(x)\, \mathrm dx = -{\ln|{\cos(x)}|} + C$ .

Proof. The immediate proof is by letting $f(x) = \cos(x)$ in Theorem 1.

Fleshing out the details, define $u = \cos(x) \Rightarrow \mathrm du = -\sin(x)\,\mathrm dx$ so that

$\displaystyle \begin{aligned} \int \tan(x)\, \mathrm dx &= \int \frac{\sin(x)}{\cos(x)}\, \mathrm dx \\ &= -\int \frac{1}{\cos(x)}\cdot -\sin(x)\, \mathrm dx \\ &= -\int \frac{1}{u}\, \mathrm du \\ &= -{\ln|u|} + C \\ &= -{\ln|{\cos(x)}|} + C. \end{aligned}$

We also obtain integrals for rational functions of the form $(\text{constant})/(\text{linear})$ and $(\text{constant})/(\text{quadratic})$ .

Technically, much more is true, but we will illustrate these simpler cases here, and expand on the “much more” in a future post.

Corollary 2. $\displaystyle \int \frac{1}{x^2 - 1}\, \mathrm dx = \frac 1{2} \ln \left| \frac{x-1}{x+1} \right| + C$ .

Proof. By writing $x^2 - 1 = (x-1)(x+1)$ and decomposing using partial fractions,

$\displaystyle \frac 1{x^2-1} = \frac 1{(x-1)(x+1)} = \frac 1{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right).$

Integrating on both sides,

$\displaystyle \begin{aligned} \int \frac 1{x^2-1}\,\mathrm dx &= \frac 1{2}\int\left(\frac{1}{x-1} - \frac{1}{x+1}\right)\, \mathrm dx \\ &= \frac 1{2}\left(\int\frac{1}{x-1}\, \mathrm dx - \int\frac{1}{x+1}\, \mathrm dx\right). \end{aligned}$

To integrate each component, we observe that $(x \pm 1)' = 1$ . Therefore, Theorem 2 yields

$\displaystyle \begin{aligned} \int \frac{1}{x \pm 1}\, \mathrm dx = \int \frac{(x \pm 1)'}{x \pm 1}\, \mathrm dx = \ln|x \pm 1| + C. \end{aligned}$

Thus, completing our integration,

$\displaystyle \begin{aligned} \int \frac 1{x^2-1}\,\mathrm dx &= \frac 1{2}\left(\int\frac{1}{x-1}\, \mathrm dx - \int\frac{1}{x+1}\, \mathrm dx\right) \\ &= \frac 1{2} \left( (\ln|x-1| + C_1) - (\ln|x-1| + C_2) \right) \\ &= \frac 1{2} \ln \left|\frac{x-1}{x+1}\right| + \frac{C_1 - C_2}{2} = \frac 1{2} \ln \left|\frac{x-1}{x+1}\right| + C.\end{aligned}$

where we abbreviated the constant $C := (C_1 - C_2)/2$ .

As a side-note, this abbreviation also explains why in each iteration of integration, we only need one $C$ , since all of the respective arbitrary constants $C_i$ coalesce into a single arbitrary constant $C$ .

Letting $u$ be a linear function also proves to be incredibly useful in integrating such functions.

Theorem 3. Let $f$ be a known differentiable function and $a,b$ be constants with $a \neq 0$ . Then

$\displaystyle \int f'(ax+b)\, \mathrm dx = \frac 1a f(ax+b) + C.$

Proof. Make the substitution $u = ax+b$ . Then the notation $\mathrm du = a\, \mathrm dx$ yields

$\displaystyle \begin{aligned} \int f'(ax+b)\, \mathrm dx &= \frac 1a \cdot \int f'(ax+b) \cdot a \, \mathrm dx \\ &= \frac 1a \int f'(u) \, \mathrm du \\ &= \frac 1a f(u) + C \\ &= \frac 1a f(ax+b) + C. \end{aligned}$

This helps us take advantage of simpler integrals to compute more complicated integrals.

Corollary 3. For any $a > 0$ , $\displaystyle \int \frac{1}{a^2 - x^2}\, \mathrm dx = \frac 1{2a} \ln \left|\frac {a-x}{a+x}\right| + C$ .

Proof. Make the substitution $x = au$ . Then the notation $\mathrm dx = a\, \mathrm du$ yields

$\displaystyle \begin{aligned} \int \frac{1}{a^2 - x^2}\, \mathrm dx &= \int \frac{1}{a^2 - (au)^2}\cdot a\, \mathrm du \\ &= a \int \frac{1}{a^2 - a^2 u^2}\, \mathrm du \\ &= -\frac {1}{a} \int \frac{1}{u^2 - 1}\, \mathrm du \\ &= -\frac 1{2a} \ln \left| \frac{u-1}{u+1}\right| + C \\ &= \frac 1{2a} \ln \left| \frac{a+x}{a-x}\right| + C,\end{aligned}$

where the final equality is left as an exercise in algebruh.

Finally, we can use known trigonometric substitutions by taking advantage of the Pythagorean identity.

Corollary 4. $\displaystyle \int \frac 1{1+x^2}\, \mathrm dx = \tan^{-1}(x) + C$ .

Proof. Make the substitution $x = \tan(u)$ . Then the notation $\mathrm dx = \sec^2(u)\, \mathrm du$ yields

$\displaystyle \begin{aligned} \int \frac{1}{1 + x^2}\, \mathrm dx &= \int \frac{1}{1 + \tan^2(u)} \cdot \sec^2(u)\, \mathrm du \\ &= \int \frac{1}{\sec^2(u)} \cdot \sec^2(u)\, \mathrm du \\ &= \int 1\, \mathrm du = u + C \\ &= \tan^{-1}(x) + C. \end{aligned}$

Sidenote: One might wonder if this could be used to derive the formula $(\tan^{-1}(x))' = 1/(1+x^2)$ . This does not work, since we needed to assume that $u = \tan^{-1}(x)$ is differentiable. If that were the case, then computation just reiterates the differentiation property.

This result, in particular, helps us integrate the reciprocal of non-factorisable quadratic functions.

Corollary 5. For any $a > 0$ , $\displaystyle \int \frac{1}{a^2 + x^2}\, \mathrm dx = \frac 1{a} \tan^{-1}\left(\frac xa\right) + C$ .

Proof. Make the substitution $x = au$ . Then the notation $\mathrm dx = a\, \mathrm du$ yields

$\displaystyle \begin{aligned} \int \frac{1}{a^2 + x^2}\, \mathrm dx &= \int \frac{1}{a^2 + (au)^2}\cdot a\, \mathrm du \\ &= \int \frac{a}{a^2(1 + u^2)}\, \mathrm du \\ &= \frac 1{a} \int \frac{1}{1 + u^2}\, \mathrm du \\ &= \frac 1{a} \tan^{-1}(u) + C \\ &= \frac 1{a} \tan^{-1}\left( \frac{x}{a} \right) + C. \end{aligned}$

Corollary 6. For any $a,b > 0$ , $\displaystyle \int \frac{1}{ a + b u^2 }\, \mathrm dx = \frac 1{\sqrt{ab}} \tan^{-1}\left(\frac {x \sqrt b}{\sqrt a}\right) + C$ .

Proof. Make the substitution $u = x\sqrt b$ . Then the notation $\displaystyle \mathrm dx = \frac 1{\sqrt b} \, \mathrm du$ yields

$\displaystyle \begin{aligned} \int \frac{1}{a + bx^2}\, \mathrm dx &= \int \frac{1}{(\sqrt a)^2 + u^2}\cdot \frac 1{\sqrt b}\, \mathrm du \\ &= \frac 1{\sqrt b} \int \frac{1}{(\sqrt a)^2 + u^2}\, \mathrm du \\ &= \frac 1{\sqrt b} \cdot \frac 1{\sqrt a} \tan^{-1}\left( \frac{u}{\sqrt a} \right) + C \\ &= \frac 1{\sqrt{ab}} \tan^{-1}\left( \frac{x\sqrt b}{\sqrt a} \right) + C. \end{aligned}$

Applications of similar techniques yields the following integral.

Theorem 4. For any $a > 0$ , $\displaystyle \int \frac{1}{\sqrt{a^2-x^2}}\, \mathrm dx = \sin^{-1}\left(\frac xa\right) + C$ .

Finally, no discussion on the reverse chain rule is complete without the notorious integral of $\sec$ .

Theorem 5. $\displaystyle \int \sec(x)\, \mathrm dx = \ln|{\sec(x) + \tan(x)}| + C$ .

Proof. By *ingenuity*, make the substitution $u = \sec(x) + \tan(x)$ . Then

$\displaystyle \frac{\mathrm du}{\mathrm dx} = (\sec(x)\tan(x) + \sec^2(x)) \quad \Rightarrow \quad \frac 1u\, \mathrm du = \sec(x)\, \mathrm dx$

yields the integration

$\begin{aligned} \displaystyle \int \sec(x)\, \mathrm dx = \int \frac 1u\, \mathrm du = \ln|u| + C = \ln|{\sec(x) + \tan(x)}| + C. \end{aligned}$

Applying similar techniques yields the following integral:

Theorem 6. For any $a > 0$ ,

$\displaystyle \int \frac{1}{\sqrt{x^2 \pm a^2}}\, \mathrm dx = \ln \left| x + \sqrt{x^2 \pm a^2} \right| + C$ .

Proof. Left as an exercise in *ingenuity*. Just kidding. But yes, an exercise in the reverse chain rule. Here are the hints:

For $\displaystyle \int \frac{1}{\sqrt{x^2 + a^2}}\, \mathrm dx$ , make the substitution $x = a \tan u$ .
For $\displaystyle \int \frac{1}{\sqrt{x^2 - a^2}}\, \mathrm dx$ , make the substitution $x = a \sec u$ .

Then invoke Theorem 5 and perform some bookkeeping via algebruh.

Next time, we reverse the product rule.

—Joel Kindiak, 25 Oct 24, 1405H

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Why u-Substitution is So Awesome

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