Curious Cheetah

Multiple u Substitutions


Calculus students often wonder how to select the “correct” \(u\) to substitute to solve an integration problem. Often, however, there are multiple acceptable candidates, any of which will result in a valid integration. This article discusses several examples of cases where different paths lead to the same result.

Problem #1



For this \(u\) substitution, we have two obvious candidates: \(\sin\theta\) and \(\cos\theta\). Let’s use the first one:
\[\begin{align} \tag{2a} u &= \sin\theta \\ \tag{2b} du &= \cos\theta\,d\theta \end{align}\]
Substituting \(u\) and \(du\) gives us:
\[\begin{align} \tag{3a} \int\sin\theta\cos\theta\,d\theta &= \int u\,du \\ \tag{3b} &= \frac{u^2}{2} + c \\ \tag{3c} &= \frac{\sin^2\theta}{2} + c \end{align} \]
Let’s do that again, but this time use the second candidate:
\[\begin{align} \tag{4a} u &= \cos\theta \\ \tag{4b} du &= -\sin\theta\,d\theta \\ \tag{4c} -du &= \sin\theta\,d\theta \end{align}\]
Substituting \(u\) and \(du\) gives us:
\[\begin{align} \tag{5a} \int\sin\theta\cos\theta\,d\theta &= -\int\cos\theta(-\sin\theta)\,d\theta \\ \tag{5b} &= -\int u\,du \\ \tag{5c} &= -\frac{u^2}{2} + c \\ \tag{5d} &= -\frac{\cos^2\theta}{2} + c \end{align}\]
Combining our answers, we get:
\[\tag{6} \int\sin\theta\cos\theta\,d\theta = \frac{\sin^2\theta}{2} + c_1 = -\frac{\cos^2\theta}{2} + c_2\]
We can demonstrate that these are in fact the same result by applying the identity \(\sin^2x+\cos^2x=1\), that is, \(\sin^2x-1=-\cos^2x\).
\[\begin{equation} \tag{7} -\frac{\cos^2\theta}{2} + c_2 = \frac{\sin^2\theta-1}{2} + c_2 = \frac{\sin^2\theta}{2} - \frac{1}{2} + c_2 \end{equation}\]
Notice that the actual value of \(c\) is flexible. Since \(-1/2\) is a constant, \(-1/2+c_2\) is likewise a constant. Hence, (3c) and (5d) are equal.

Problem #2

\[\tag{8} \int_{0}^{1}(x+1)(x^2+2x+1)\,dx \]


There are several approaches we could use for this problem. We could multiply the polynomials and integrate the result, use \(u\) substitution on the second term, or factor the second term and use a different \(u\) substitution. Let’s start with the first approach:
\[\begin{align} \tag{9a} (x+1)(x^2+2x+1) &= (x^3+2x^2+x)+(x^2+2x+1) \\ \tag{9b} &= x^3 + 3x^2 + 3x + 1 \\ \tag{9c} \int_0^1(x+1)(x^2+2x+1)\,dx &= \int_0^1 x^3 + 3x^2 + 3x + 1 \,dx \\ \tag{9d} &= \left. \frac{x^4}{4} + 3\cdot\frac{x^3}{3} + 3\cdot\frac{x^2}{2} + {x} \,\right|_0^1 \\ \tag{9e} &= \left(\frac{1^4}{4} + 1^3 + \frac{3\cdot 1^2}{2} + 1\right) - \left(\frac{0^4}{4} + 0^3 + \frac{3\cdot 0^2}{2} + 0\right) \\ \tag{9f} \label{x3sol1} &= \frac{1}{4} + 1 + \frac{3}{2} + 1 = 3\frac{3}{4} \end{align}\]
Now let’s try the second approach, where \(u\) is the second term of (8):
\[\begin{align} \tag{10a} u &= x^2+2x+1 \\ \tag{10b} du &= (2x+2)\,dx \\ \tag{10c} &= 2(x+1)\,dx \\ \tag{10d} \label{x3sol2du} \frac{du}{2} &= (x+1)\,dx \end{align}\]
Substituting gives us:
\begin{align} \tag{11a} \int_0^1(x+1)(x^2+2x+1)\,dx &= \frac{1}{2}\int_{x=0}^{x=1} u\,dx \\ \tag{11b} &= \left. \frac{1}{2} \cdot \frac{u^2}{2} \,\right|_{x=0}^{x=1} \\ \tag{11c} &= \left. \frac{u^2}{4} \,\right|_{1}^{4} & \text{(10d)} \\ \tag{11d} \label{x3sol2} &= \frac{16}{4} - \frac{1}{4} = 3\frac{3}{4} \end{align}
Finally, notice that we can factor \(x^2+2x+1\) to \((x+1)^2\). We can replace this in (8):
\[\begin{equation} \tag{12} \int_0^1(x+1)(x^2+2x+1)\,dx = \int_0^1(x+1)^3\,dx \end{equation}\]
The \(u\) substitution this time is easier:
\[\begin{align} \tag{13a} u &= x+1 \\ \tag{13b} du &= dx \end{align}\]
which produces:
\[\begin{align} \tag{14a} \int_0^1(x+1)^3\,dx &= \int_{x=0}^{x=1} u^3\,du \\ \tag{14b} &= \left. \frac{u^4}{4} \,\right|_{x=0}^{x=1} \\ \tag{14c} &= \left. \frac{u^4}{4} \,\right|_1^2 & \text{(13b)} \\ \tag{14d} &= \frac{16}{4} - \frac{1}{4} = 3\frac{3}{4} \end{align}\]
As expected, (9f), (11d), and (14d) have the same value. Regardless of which method we use, we should always get the same result.

Problem #3

\[\begin{equation}\tag{15}\int 8x\cot x^2\csc x^2\,dx\end{equation}\]


There are at least three ways to solve this, depending on what we choose as \(u\). The options we'll look at are \(x^2\), \(\csc x^2\), and \(\sin x^2\). Again, let’s start with the first:
\[\begin{align} \tag{16a} u &= x^2 \\ \tag{16b} du &= 2x\,dx \end{align} \]
which produces:
\[\begin{align} \tag{17a} \int 8x\cot x^2\csc x^2\,dx &= 4\int \cot u\csc u\,du \\ \tag{17b} &= -4\int-\cot u\csc u\,du \\ \tag{17c} &= -4\csc u + c \\ \tag{17d} &= -4\csc x^2 + c \end{align} \]
The second method relies on the same basic derivation, i.e., \(d(\csc\theta)=\csc\theta\cot\theta\,d\theta\).
\[\begin{align} \tag{18a} u &= \csc x^2 \\ \tag{18b} du &= -\csc x^2 \cot x^2 \,d(x^2) \\ \tag{18c} &= -2x\csc x^2 \cot x^2 \,dx \end{align} \]
Note that we only need to substitute for \(du\), rather than both \(u\) and \(du\):
\[\begin{align} \tag{19a} \int 8x\cot x^2\csc x^2\,dx &= \int -4(-2x\cot x^2\csc x^2)\,dx \\ \tag{19b} &= -4\int\,du \\ \tag{19c} &= -4(u + c) \\ \tag{19d} &= -4\csc x^2 + c \end{align} \]
For the third method, we need to convert the trigonometric functions into the primitives:
\[\begin{align}\tag{20a}\cot\theta&=\frac{\cos\theta}{\sin\theta} \\ \tag{20b}\csc\theta &= \frac{1}{\sin\theta} \\ \tag{20c} \cot\theta\csc\theta &= \frac{1\cdot\cos\theta}{\sin\theta\cdot\sin\theta} \\ \tag{20d} &= \frac{\cos\theta}{\sin^2\theta} \end{align} \]
We will be using the following \(u\) substitution:
\[\begin{align} \tag{21a} u &= \sin x^2 \\ \tag{21b} du &= \cos x^2\,d(x^2) \\ \tag{21c} &= 2x\cos x^2\,dx \end{align}\]
We are now prepared to do the substition:
\[\begin{align} \tag{22a} \int 8x\cot x^2\csc x^2\,dx &= \int 8x\frac{\cos x^2}{\sin^2 x^2}\,dx & \text{(20d)} \\ \tag{22b} \int 4\frac{2x\cos x^2\,dx}{\sin^2x^2} \\ \tag{22c} &= 4\int\frac{du}{u^2} & \text{(21a, 21c)}\\ \tag{22d} &= 4\int u^{-2}\,du \\ \tag{22e} &= 4(-u^{-1} + c) \\ \tag{22f} &= \frac{-4}{u} + c \\ \tag{22g} &= \frac{-4}{\sin x^2} + c \\ \tag{22h} &= -4\csc x^2 + c \end{align} \]
As expected, (17d), (19d), and (22h) are the same.
In each of these examples, we have seen that we can start with different substitutions and end up with substantively identical substitutions. The art and mastery of \(u\) substitution lies in developing a sense of the most efficient substitution to make, but in many cases there’s not a single “correct” substitution.