Reverse Chain Rule

Section 6.1 Reverse Chain Rule

With integration, a common situation is when you recognize a basic function you know how to integrate, except the input \(x\) has been replaced by something like \(3x\text{,}\) \(2x+5\text{,}\) or \(1-4x\) (a linear expression). In these cases, the antiderivative has the same shape as usual, but you must make a small adjustment.

🔗

Example 6.1.1. Integral of \(e^{3x}\).

Consider the integral,

🔗

\begin{equation*} \int e^{3x} \,dx \end{equation*}

Solution.

We know that the integral of \(e^x\) is \(e^x + C\text{,}\) so we might guess that the integral of \(e^{3x}\) is \(e^{3x} + C\text{.}\) However, this is almost correct but not quite. Indeed, checking the antiderivative,

🔗

\begin{equation*} \frac{d}{dx} (e^{3x}) = 3e^{3x} \qquad \text{which is not } e^{3x} \end{equation*}

There is an extra factor of 3. When differentiating \(e^{3x}\text{,}\) we know that we’d basically differentiate as normal, and then multiply by the derivative of \(3x\) which is 3 (by the chain rule), and so the derivative is \(3e^{3x}\text{.}\) Antiderivatives are like the reverse of derivatives, so instead, we need to divide by 3,

🔗

\begin{equation*} \int e^{3x} \,dx = \frac{1}{3} e^{3x} + C \end{equation*}

Checking the antiderivative, indeed,

🔗

\begin{align*} \frac{d}{dx} \brac{\frac{1}{3} e^{3x}} \amp = \frac{1}{3} \cdot 3e^{3x}\\ \amp = e^{3x} \end{align*}

which is the original function.

🔗

Example 6.1.2. Integral of \(\sin{(2x + 5)}\).

Consider the integral,

🔗

\begin{equation*} \int \sin{(2x + 5)} \,dx \end{equation*}

Solution.

Again, we know the integral of \(\sin{x}\) is \(-\cos{x} + C\text{,}\) so we might guess that the integral of \(\sin{(2x + 5)}\) is \(-\cos{(2x + 5)} + C\text{.}\) However, this is almost correct but not quite. We also have to divide the answer by 2, because differentiating would make an extra factor of 2 (from the chain rule). Then,

🔗

\begin{equation*} \int \sin{(2x + 5)} \,dx = -\frac{1}{2}\cos{(2x + 5)} + C \end{equation*}

Checking the antiderivative verifies that this is correct,

🔗

\begin{align*} \frac{d}{dx} \brac{-\frac{1}{2}\cos{(2x + 5)}} \amp = -\frac{1}{2} \cdot -2 \sin{(2x + 5)}\\ \amp = \sin{(2x + 5)} \end{align*}

which gives back the original function. Notice that the \(+5\) does not change the answer, because the derivative of a constant is 0.

🔗

In general, if there is a linear expression \(ax + b\) inside the function we are integrating, then we need to divide the answer by \(a\) to correct for that extra constant factor when differentiating. More precisely, differentiating something with \(ax+b\) inside always leads to a factor of \(a\) by the chain rule, so when we integrate we “undo” that factor by dividing by \(a\text{.}\) The \(+b\) does not affect the factor because the derivative of a constant is 0. In short,

🔗

\begin{equation*} \boxed{\text{If there's just } ax+b \text{ instead of } x \text{, then integrate like normal and put } \frac{1}{a} \text{ in front}} \end{equation*}

Example 6.1.3. Integral of \(\sec^2(2x)\).

Evaluate \(\int \sec^2(2x) \,dx\text{.}\)

🔗

Solution.

The antiderivative of \(\sec^2(x)\) is \(\tan(x)\text{.}\) Here we have \(2x\) instead of \(x\text{,}\) so we integrate as normal and divide by the coefficient of \(x\text{,}\) which is 2,

🔗

\(\int \sec 2x\tan 2x \,dx\)

🔗

Answer.

\(\frac{1}{2}\sec 2x+C\)

🔗

The same idea applies to powers and logarithms.

🔗

Example 6.1.4. Integral of \((1-2x)^9\).

Evaluate \(\int (1-2x)^9 \,dx\text{.}\)

🔗

Solution.

This integral is of the form \(u^9\) (a power), except with \(1-2x\) inside (a linear expression). So, we integrate \((1-2x)^9\) as if it were \(u^9\text{,}\) giving \(\frac{1}{10}u^{10}\) (by the power rule), and then divide by the coefficient of \(x\text{,}\) which is \(-2\text{,}\)

🔗

\begin{align*} \int (1-2x)^9 \,dx \amp = \frac{1}{-2} \cdot \frac{(1-2x)^{10}}{10} + C\\ \amp = -\frac{1}{20}(1-2x)^{10} + C \end{align*}

🔗

Example 6.1.5. Integral of \(\frac{1}{5-3x}\).

Evaluate \(\int \frac{1}{5-3x} \,dx\text{.}\)

🔗

Solution.

This integral is of the form \(\frac{1}{u}\text{,}\) except with \(5-3x\) (a linear expression). The antiderivative of \(\frac{1}{u}\) is \(\ln\abs{u}\text{,}\) so we integrate as normal and divide by the coefficient of \(x\text{,}\) which is \(-3\text{,}\)

🔗

\begin{align*} \int \frac{dx}{5-3x} \amp = \frac{1}{-3} \ln\abs{5-3x} + C\\ \amp = -\frac{1}{3}\ln\abs{5-3x} + C \end{align*}

🔗

Example 6.1.6. Integral of \(\sqrt{1+4x}\).

Evaluate \(\int \sqrt{1+4x} \,dx\text{.}\)

🔗

Solution.

This integral is of the form \(\sqrt{u}\text{,}\) except with \(1+4x\) (a linear expression). The antiderivative of \(\sqrt{u}\) is \(\frac{2}{3}u^{3/2}\text{,}\) so we integrate as normal and divide by the coefficient of \(x\text{,}\) which is 4,

🔗

\(\int \sqrt{x-1}\,dx\)

🔗

Answer.

\(\frac{2}{3}(x-1)^{3/2}+C\)

🔗

In general, we can summarize this rule,

🔗

Theorem 6.1.7. Reverse Chain Rule (Linear Inside Rule).

If \(\int f(x) \,dx = F(x)\text{,}\) then \(\int f(ax + b) \,dx = \frac{1}{a} F(ax + b)\text{.}\)

🔗

Here, \(\int f(x) \,dx = F(x)\) means that there is some existing integral rule that we know for \(f(x)\text{,}\) and \(F(x)\) is the antiderivative of \(f(x)\text{.}\) Then, if we have \(f(ax + b)\) instead of \(f(x)\) (in other words, \(ax+b\) instead of just \(x\)), we can integrate as normal to get \(F(ax + b)\text{,}\) and then divide by \(a\) to correct for the chain rule. For example,

🔗

\begin{equation*} \int e^{ax + b} \,dx = \frac{1}{a} e^{ax + b} + C \end{equation*}

Remark 6.1.8.

This rule only works when the inner function is linear (in other words, of the form \(ax+b\)), because the derivative of a linear function is a constant. For an inner function that isn’t linear, this shortcut does not apply. For example, for \(\int \cos{(x^2)} \,dx\text{,}\) we cannot just integrate as normal and divide by the derivative of \(x^2\) which is \(2x\) to get \(\frac{1}{2x}\sin{(x^2)} + C\text{,}\)

🔗

\begin{equation*} \int \cos{(x^2)} \,dx \neq \frac{1}{2x}\sin{(x^2)} + C \end{equation*}

This is because when we differentiate \(\frac{1}{2x}\sin{(x^2)} + C\text{,}\) we have to use the product rule, and we will get something more complicated than just \(\cos{(x^2)}\text{,}\) so this is not the correct antiderivative. We will learn how to integrate some more complicated functions later.

🔗