Suppose we want to evaluate an integral of the form \[\int f(g(x))g'(x)dx\tag{a}\] To this end, let \(g(x)=u.\) Finding the differential of both sides, we have
\[g'(x)dx=du.\tag{b}\] If we substitute (b) into (a), we will have \[\bbox[#F2F2F2,5px,border:2px solid black]{\int f(\underbrace{g(x)}_{u})\underbrace{g'(x)dx}_{du}=\int f(u)du.\tag{c}}\] This equation supplies us with a method for determining integrals in a large number of cases in which the form of the integral is not obvious.
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Because \(F’=f\) it follows from the chain rule that \[\begin{aligned} \frac{d}{dx}F(g(x)) & =F'(g(x))g'(x)\\ & =f(g(x))g'(x).\end{aligned}\] It follows from the definition of antiderivative (or integral) that \[\begin{align*} \int f(g(x))g'(x)dx & =F(g(x))+C\\ & =F(u)+C\tag{${u=g(x)}$}\\ & =\int F'(u)du\\ & =\int f(u)du. \end{align*}\]
- Because traditionally the letter \(u\) is used in the Substitution Rule, it is sometimes called \(u\)-substituiton, but instead of \(u\), we can use any letter such as \(t,w,\theta,\) etc.
Formula (c) may be stated as an algorithm as follows
- Choose \(u\), say put \(g(x)=u\).
- Compute \(du=g'(x)dx\).
- Substitute \(u\) and \(du\) in the integral. At this step, everything should be expressed in terms of \(u\) and no \(x\) should be present in the integral. If it is not possible, make another choice for \(u\).
- Evaluate the resulting integral.
- Replace \(u\) by \(g(x)\) and express the final result in terms of \(x\).
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Let \(u=ax+b\). Then \(du=a\ dx\) or \(dx=\frac{1}{a}du\) and \[\begin{align*} \int f(ax+b)dx & =\int\frac{1}{a}f(u)\ du\\ & =\frac{1}{a}F(u)+C\\ & =\frac{1}{a}F(ax+b)+C\tag{${u=ax+b}$} \end{align*}\]
\[\bbox[#F2F2F2,5px,border:2px solid black]{\int\tan x\ dx=\ln|\sec x|+C\quad\text{or}\quad-\ln|\cos x|+C}\]
\[\bbox[#F2F2F2,5px,border:2px solid black]{\begin{array}{c} {\displaystyle \int\sinh x\ dx=\cosh x+C}\\ {\displaystyle \int\cosh x\ dx=\sinh x+C} \end{array}}\]