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Section 4.2 Integration by Substitution

First, let's go over a few corrolations. Since derivatives and integrals are inverses of each other. The following operations are inverses as well:

Table 4.2.1.
RULES
power \(\longrightarrow\) power
trig \(\longrightarrow\) trig
quotient \(\longrightarrow\) N/A
chain \(\longrightarrow\) u-substitution
RECALL.
\begin{align*} \dfrac{d}{dx}\left( (1+ 3x^2)^7 \right) \amp = 7(1 + 3x^2)^6 \cdot 6x\\ \amp = 42x (1 + 3x^2)^6 \end{align*}

IDEA:

\(\)
\begin{equation*} \nearrow \dfrac{d}{dx} (\cdots) \searrow \end{equation*}
\(\)
\(F(x) = (1 + 3x^2)^7\)
\begin{equation*} \rightleftharpoons \end{equation*}
\(f(x) = 42x(1 + 3x^2)^7\)
\(\)
\begin{equation*} \nwarrow \int \cdots dx \swarrow \end{equation*}
\(\)

Example 4.2.1.

Evaluate \(\displaystyle\int 42x (1 + 3x^2)^6 dx\) by u-substitution:

  1. choose a \(u = g(x)\) that matches the "inside function"

    \begin{equation*} u = g(x) = 1 + 3x^2 \end{equation*}
  2. define differential \(du\text{;}\) recall \(du = g'(x) dx\)

    \begin{equation*} du = 6x dx \end{equation*}
  3. rewrite original integral in terms of \(u\) and \(du\)

    \begin{align*} \int 42x ( 1 + 3x^2)^6 dx \amp = \int \underbrace{(1 + 3x^2)^6}_{u^6} \cdot \underbrace{42x dx}_{7du}\\ \amp = \int 7u^6 du\\ \amp = 7 \left( \dfrac{u^7}{7} \right) + c \\ \amp = (1 + 3x^2)^7 + c \end{align*}