U-substitution

We've already learned about finding direct antiderivatives to common functions. However, integrands seldom consist of pure unadulterated functions in this manner. In this topic, we will take a look at the relationship between the chain rule and the integration of composite functions. Moreover, we'll learn about a technique used to reshape and simplify these integrands into expressions that have known direct antiderivatives.

Integration of composite functions

Let's consider the following derivative

$$\frac{d}{dx} f(g(x))$$Using the chain rule, we know that

$$\frac{d}{dx} f(g(x)) = \frac{d}{dg} f(x) \cdot \frac{d}{dx} g(x)$$Thus, the following must be true

$$\int \frac{d}{dg} f(x) \cdot \frac{d}{dx} g(x) dx = f(g(x)) + C$$As an example, let's consider the following integral:

$$\int \tan(x) dx$$ First, let's recall that

$$\tan(x) = \frac{\sin(x)}{\cos(x)}$$Then, we can rewrite our original integral as

$$\int \frac{1}{\cos(x)} \cdot \sin(x) dx$$Let's also recall the antiderivative of $\frac{1}{x}$,

$$\frac{d}{dx} \ln(x) = \frac{1}{x} \implies \int \frac{1}{x} dx = \ln|x| + C$$We can generalize the argument of the logarithm as $g(x)$, and apply the chain rule as follows

$$\frac{d}{dx} \ln(g(x)) = \frac{1}{g(x)} \cdot\frac{d}{dx} g(x) \implies \int \frac{1}{g(x)} \cdot g'(x) dx = \ln|g(x)| + C$$Then, if we define

$$g(x) = \cos(x)$$we'll have the following

$$\int \frac{1}{\cos(x)} \cdot (\cos(x))' dx = \ln|\cos(x)| + C \Rightarrow -\int \frac{1}{\cos(x)} \cdot \sin(x) dx = -\ln|\cos(x)| + C$$Or, in other words

$$\int \tan(x) dx = -\ln|\cos(x)| + C$$

U-substitution

In the previous example, while trying to solve the integral $\int \tan(x) dx$, we were able to integrate the following composite function by applying the chain rule:

$$f(g(x)) = \frac{1}{\cos(x)} \quad ; \quad f(x)=\frac{1}{x} \text{ and } g(x) = \cos(x)$$since we knew the derivative of the inner function $g'(x) = -\sin(x)$ was the remaining factor of the integrand.

In order to find the right combination of factors to fit this reverse chain rule process, a technique called u-substitution (or also integration by change of variables) is often employed. Let's take a look at how it works with another example:

$$I=\int 2x \cdot 3^{x^2+5} \ dx$$Here, $I$ is a placeholder for the antiderivative resulting from solving the integral.

Let's define the following substitution:

$$u = x^2 + 5$$We can differentiate both sides to obtain

$$\frac{d}{dx}u = \frac{d}{dx}(x^2 + 5) \implies \frac{du}{dx} = 2x \implies du = 2xdx$$Substituting our new expressions in the original integral,

$$\begin{align*} I &= \int 2x \cdot 3^{x^2+5} \ dx = \int 3^{x^2+5} \cdot 2x \ dx \\ I &= \int 3^u du \end{align*}$$Now, we already know the direct antiderivative of that expression!

$$\int 3^u du = \frac{3^u}{\ln(3)} + C$$All we need to do now is substitute back our expression for $u$ to find the solution of the original integral:

$$I = \frac{3^{(x^2 + 5)}}{\ln(3)} + C$$

Let's try u-substitution with the same example from the previous section:

$$I=\int \tan(x) dx = \int \frac{\sin(x)}{\cos(x)}dx$$Let's define $u$,

$$u = \cos(x)$$And find $du$

$$\frac{d}{dx} u = \frac{d}{dx}\cos(x) \implies \frac{du}{dx} = -\sin(x) \implies du = -\sin(x)dx \implies -du = \sin(x)dx$$

Then, we have

$$I = -\int \frac{1}{u}du$$We already know how to solve that integral:

$$I=-\int \frac{1}{u}du = -\ln|u|+C$$Now, we reverse the change of variables $u=\cos(x)$

$$I= -\ln|\cos(x)|+C$$Finally,

$$\int \tan(x) dx = -\ln|\cos(x)|+C$$

U-substitution: integral of the secant function

We can use $u$-substitution to find the antiderivative for the secant function:

$$I =\int \sec(x) dx$$First, let's recall a couple of derivatives:

$$\begin{rcases} &\frac{d}{dx} \sec(x) = \sec(x) \cdot\tan(x) \\ &\frac{d}{dx} \tan(x) = \sec^2(x) \end{rcases} \implies \frac{d}{dx} \left(\sec(x) + \tan(x) \right) = \sec(x) \cdot \left( \tan(x) + \sec(x) \right)$$With this knowledge, let's manipulate our integrand by multiplying and dividing by $\left( \tan(x) + \sec(x) \right)$, like so

$$\int \sec(x) \ dx = \int \sec(x) \cdot \left( \frac{\tan(x) + \sec(x)}{\tan(x) + \sec(x)} \right) dx$$Now, we can define $u$

$$u = \tan(x) + \sec(x) \implies du = \sec(x) \cdot \left( \tan(x) + \sec(x) \right) \ dx$$Substituting in,

$$I = \int \sec(x) \cdot \left( \frac{\tan(x) + \sec(x)}{\tan(x) + \sec(x)} \right) dx \implies I = \int \frac{1}{u} \ du$$We already know the antiderivative for this integral,

$$I = \int \frac{1}{u} \ du = \ln|u| + C$$So, we only need to reverse our change of variables

$$I = \ln|u| + C \implies I = \ln|\tan(x) + \sec(x) | +C$$Finally,

$$\int \sec(x) dx = \ln|\tan(x) + \sec(x) | +C$$

U-substitution: another example

Let's consider the following integral

$$I =\int \frac{1}{(x+1) \cdot x} \ dx$$

We can perform some algebraic manipulation to break down this integral before doing any change of variables,

$$\begin{align*} I &= \int \frac{1}{(x - 1) \cdot x} \ dx = \int \frac{x-x+1}{(x - 1) \cdot x} \ dx = -\int \frac{-x + x -1}{(x -1) \cdot x} \ dx \\ I &= - \left( \int \frac{-x}{(x - 1) \cdot x} \ dx + \int \frac{x-1}{(x - 1) \cdot x} \ dx \right) \\ I &= \int \frac{1}{x - 1} \ dx - \int \frac{1}{x} \ dx \end{align*}$$Or, in other words

$$I = I_1 + I_2 \quad ; \text{ with } I_1 =\int \frac{1}{x - 1} \ dx \text{ and } I_2 =- \int \frac{1}{x} \ dx$$

We know $I_2$ has a direct antiderivative,

$$I_2 = - \ln|x| +C$$So, let's perform a $u$-substitution on $I_1$ to find its antiderivative:

$$\begin{align*} u = x-1 \implies \frac{d}{dx}u &= \frac{d}{dx}(x-1) \\ du &= 1 \cdot dx \\ du &= dx \end{align*}$$Then,

$$I_1 = \int \frac{1}{x-1} \ dx = \int \frac{1}{u} \ du = \ln|u| + C$$Reversing the change of variables,

$$I_1 = \ln|u| + C \implies I_1 = \ln|x-1| + C$$Thus,

$$I =\ln|x-1| - \ln|x| + C$$

Two approaches to the u-substitution method

Let's consider the following integral

$$I = \int \frac{1}{e^x + 1} dx$$There are two ways to approach a change of variables: either to define the $u$-substitution and differentiate implicitly to find $du$, or to define the $u$-substitution, solve for $x$ and then differentiate. Let's take a look at both.

First approach: differentiate implicitly to find $du$

Let's start by defining the $u$-substitution:

$$u = e^x + 1$$Now, let's define $du$:

$$\frac{d}{dx} u = \frac{d}{dx} (e^x +1) \implies du = e^x dx$$

In this case, we can see the expression $e^x dx$ does not appear anywhere in the original integral. To change this, let's multiply and divide the original integrand by $e^x$:

$$I = \int \frac{e^x}{e^x} \cdot \frac{1}{e^x + 1} dx = \int \frac{e^x}{e^x \cdot (e^x + 1)} \ dx$$Now, because $u = e^x + 1$, then

$$e^x = u - 1$$

Substituting in our new expressions,$$I = \int \frac{e^x}{e^x \cdot (e^x + 1)} \ dx = \int \frac{1}{(u - 1) \cdot u} \ du$$We already know the solution for this integral from the previous section:

$$I =\ln|u-1| - \ln|u| + C$$So, we just perform the change of variables back to obtain

$$I =\ln(e^x) - \ln(e^x+1) + C$$

Second approach: solve for $x$ and then differentiate

First, let's define the $u$-substitution:

$$u = e^x + 1$$Now, let's solve for $x$,

$$x = \ln(u -1)$$Differentiate both sides to find $dx$,

$$\frac{d}{du} x = \frac{d}{du}\ln(u-1) \implies \frac{dx}{du} =\frac{1}{u -1} \implies dx = \frac{1}{u -1} \ du$$Substituting both $x$ and $dx$ in the original integral,

$$I = \int \frac{1}{e^x + 1} dx \implies I = \int \frac{1}{u} \cdot \frac{1}{u - 1} \ du = \int \frac{1}{u \cdot (u - 1)} \ du$$Which is the same integral as before. Then,

$$I = \int \frac{1}{u \cdot (u - 1)} dx = \ln|u-1| - \ln|u| + C$$And,

$$I = \ln(e^x) - \ln(e^x + 1) + C$$

Change of variables: yet another example

$$I=\int \frac{5}{4+\sqrt{x+3}} \ dx$$Let's define the following substitution:

$$u = \sqrt{x+3}$$Solving for $x$,

$$x=u^2 -3$$Differentiating both sides to find $dx$,

$$\frac{dx}{du} = \frac{d}{du} (u^2-3) \implies \frac{dx}{du} = 2u \implies dx = 2u \ du$$

Substitute our new expressions for $x$ and $dx$ in the original integral,

$$I = \int \frac{5}{4+u} \cdot 2u \ du$$

Here, we need to perform yet another change of variables. Let's use the symbol $v$ for this one

$$v = 4 + u$$Then,

$$\frac{d}{du}v = \frac{d}{du} (4+u) \implies dv = du$$And so,

$$I = \int \frac{5}{4+u} \cdot 2u \ du \implies I = \int \frac{5}{v} \cdot 2(v-4) \ dv$$Rearranging and taking constants out of the integral,

$$I = 10 \cdot \int \frac{v-4}{v} \ dv = 10 \cdot \left( \int dv - 4 \int \frac{1}{v} \ dv \right)$$We are left with two integrals with direct antiderivatives:

$$I = 10 \int dv - 40 \int \frac{1}{v} \ dv = 10 v - 40\ln|v| +C$$Finally, we reverse all the changes of the variable from the last one to the first one:

$$I = 10 v - 40\ln|v| +C \implies I = 10 \cdot (4 + u) - 40 \ln|4+u| +C \implies 10 \cdot (4 + \sqrt{x+3}) - 40 \ln\left|4+ \sqrt{x+3}\right| +C$$Thus,

$$\int \frac{5}{4+\sqrt{x+3}} \ dx = 40 + 10\sqrt{x+3} - 40 \ln \left|4+\sqrt{x+3} \right| +C$$

Common pitfalls of u-substitution

When defining $u$ we choose an expression that is a part of the original integrand. The resulting integral, after substituting both $u$ and $du$must be easier to solve than the original one; in fact, ideally it should be an integral with a direct antiderivative. If the resulting integral is more difficult to solve than the original one, you might want to redefine $u$ and try again.

Remember that the antiderivative of a composite function is not just the antiderivative of the outer-most function. Not taking into account the remaining factors when defining $u$ can lead to unexpected results. It is always good to remember that $u$-substitution is all about reversing the chain rule. So the choice of factors that define $u$ must be such that the new integrands are of the form: $f(u) \cdot u' du$

For example, let's consider the following integral

$$I=\int e^{3x^2+2} \cdot (2x) \ dx$$

We might be tempted to define $u$ as follows

$$u = 2x \implies x = \frac{1}{2} u \implies dx = \frac{1}{2} du$$But then, we'll have

$$I=\frac{1}{2}\int e^{\frac{3}{4}u^2+2} \cdot u \ du$$Which is hardly a simpler integral.

However, if we remember that

$$\frac{d}{dx} (3x^2+2) = 6x \ dx$$We know we have a similar factor in the same integrand.

Thus, if instead, we define $u$ as

$$u = 3x^2 +2 \implies du = 6x \ dx$$We'll have

$$I =\frac{1}{3}\int e^u \ du$$Which is a much simpler integral to solve!

Conclusion

In this topic we have learned:

• Integrals that have composite functions as antiderivatives can be solved by applying the chain rule in reverse.
• The $u$-substitution technique can help us with this process by simplifying more complicated expressions into simpler ones, usually with direct antiderivatives.
• The $u$-substitution technique can be applied as many times as necessary, as long as all the changes of variable are reversed.
• There are two approaches after having defined the $u$-substitution itself: differentiate implicitly to find $du$, or solve for $x$ and then differentiate to find $dx$.

In addition, we've learned the rule to determine the antiderivative for the secant function:

$$f(x) = \sec(x) \implies\int \sec(x) dx = \ln|\tan(x) + \sec(x) | +C$$