U-Substitution of Indefinite Integrals

U-Substitution of Indefinite Integrals

We will now introduce one of many methods to integration that will allow us to integrate more difficult functions. This method is known as U-Substitution or more generally as Substitution.

Suppose we have a function in the form of $f(x) = g(x)g'(x)$, and we want to find the indefinite integral of $f$, that is we want to evaluate the integral $\int f(x) \: dx$.

For example, let's say we have the function $f(x) = \sin x \cos x$ and we want to calculate $\int \sin x \cos x \: dx$. Clearly we can see that $f$ is in the form $f(x) = g(x)g'(x)$, since if $g(x) = \sin x$, then $g'(x) = \cos x$. First, let's define a variable, usually $u$, to be $u = \sin x$. If we differentiate $u$ with respect to $x$, we get that $\frac{du}{dx} = \cos x$. Multiplying both sides by the differential $dx$, we get $du = \cos x \: dx$

Now notice that our integral is $\int \sin x \cos x \: dx$. We can now substitute $u = \sin x$, and $du = \cos x \: dx$, and therefore:

(1)
\begin{align} \int \sin x \cos x \: dx = \int u \: du \end{align}

The integral on the right is much easier to evaluate, in fact, $\int u \: du = \frac{u^2}{2} + C$. We now substitute back in $u = \sin x$ to get that $\int \sin x \cos x \: dx = \frac{\sin ^2 x }{2} + C$.

Example 1

Find the antiderivative of the function $f(x) = (2 + \sin x)^3 \cos x \:$.

So we want to determine $\int (2 - \sin x)^3 \cos x \: dx$. We will let $u = 2 + \sin x$ and therefore, $du = \cos x \: dx$. Making the appropriate substitution, we get that:

(2)
\begin{align} \int (2 - \sin x)^3 \cos x \: dx = \int u^3 \: du \end{align}

When evaluated and simplified in the following manner, we obtain:

(3)
\begin{align} \int u^3 \: du = \frac{u^4}{4} = \frac{(2 + \sin x)^4}{4} \end{align}

Example 2

Find the antiderivative of the function $f(x) = (3x^2 + 3x + 2)^5 (2x + 1) \:$.

So we want to evaluate $\int (3x^2 + 3x + 2)^5 (2x + 1) \: dx$. First let $u = 3x^2 + 3x$ and $du = 6x + 3$, and let's factor our a 3 to get $du = 3(2x + 1)$.

We cannot substitute this in our expression though, since we have the term $2x + 1$ in our function $f$ and $du = 3(2x + 1)$. Instead, we divide both sides by 3 to get $\frac{du}{3} = (2x + 1) \: dx$. Now we can make the appropriate substitution:

(4)
\begin{align} \int (3x^2 + 3x + 2)^5 (2x + 1) \: dx = \int \frac{1}{3} u^5 \: du = \frac{1}{3} \int u^5 \: du \\ \end{align}

And evaluate the integral easily:

(5)
\begin{align} \frac{1}{3} \int u^5 \: du = \frac{1}{3} \frac{u^6}{6} \\ = \frac{(3x^2 + 3x + 2)^6}{18} \end{align}

Proof of the Indefinite Integral of Tangent

We now have the ability to determine the indefinite integral of $f(x) = \tan x$.

 Theorem 1: If $f(x) = \tan x$ then $\int \tan x \: dx = \ln \mid \sec x \mid + C$.
• Proof: Recall that $\tan x \frac{\sin x}{\cos x}$. Thus it follows that:
(6)
\begin{align} \int \tan x \: dx = \int \frac{\sin x}{\cos x} \: dx \end{align}
• Now let $u = \cos x$, and thus $du = -\sin x dx$. Let's make our substitutions:
(7)
\begin{align} \int \frac{\sin x}{\cos x} \: dx \\ = \int -\frac{1}{u} \: du \\ = -\ln u + C \\ = -\ln (\cos x) + C \\ = \ln ((\cos x)^{-1}) + C \\ = \ln (\sec x) + C \\ = \ln \mid \sec x \mid + C \end{align}
• We note that since $\sec x < 0$ for some values of $x$, and $\ln x$ has a domain of $(0, \infty)$, we must thus apply the absolute value to secant. $\blacksquare$