a lecture on integration by substitution

The Method of Substitution (or Change of Variables)

The following is a general method to find indefinite integrals that look like the result of a chain rule.

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When to use it: We use the method of substitution for indefinite integrals which look like the result of a chain rule. In particular, try to use this method when you see a composition of two functions.
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How to use it: In this method, we go from integrating with respect to $x$ to integrating with respect to a new variable, $u$ , which makes the integral much easier.
1. (a)
  
  Find inside the integral the composition of two functions and set $u=$ “the inner function”.
2. (b)
  
  We also write $du=\frac{du}{dx}dx$ .
3. (c)
  
  Substitute everything in the integral that depends on $x$ in terms of $u$ .
4. (d)
  
  Integrate with respect to $u$ .
5. (e)
  
  Once we have the result of integration in terms of $u$ ( $+C$ ), substitute back in terms of $x$ .

The method is best explained through examples:

Example 0.1.

We want to find $\int e^{2x}dx$ . The integrand is $e^{2x}$ , which is a composition of two functions. The inner function is $2x$ so we set:

$u=2x,\quad du=2dx$

Thus,

$x=u/2,\quad dx=du/2$

Substitute into the integral:

$\int e^{2x}dx=\int e^{u}\frac{du}{2}=\frac{1}{2}\int e^{u}du=\frac{1}{2}e^{u}+% C=\frac{1}{2}e^{2x}+C$

The following are typical examples where we use the subsitution method:

Example 0.2.

$\int xe^{3x^{2}+7}dx$

The inner function is $u=3x^{2}+7$ and $du=6xdx$ . Thus $dx=du/(6x)$ . Substitute:

$\int xe^{3x^{2}+7}dx=\int\frac{xe^{u}}{6x}du=\int\frac{e^{u}}{6}du=\frac{e^{u}% }{6}+C=\frac{e^{3x^{2}+7}}{6}+C.$

Example 0.3.

$\int\sin(3x+7)dx$

The inner function is $u=3x+7$ and $du=3dx$ . Therefore:

$\int\sin(3x+7)dx=\int\frac{\sin(u)}{3}du=-\frac{\cos(u)}{3}+C=-\frac{\cos(3x+7% )}{3}+C.$

Example 0.4.

$\int(2x+3)\sqrt{x^{2}+3x+20}\ dx$

Inner $u=x^{2}+3x+20$ and $du=(2x+3)dx$ . Thus:

$\int(2x+3)\sqrt{x^{2}+3x+20}\ dx=\int\sqrt{u}du=\int u^{1/2}du=\frac{2u^{3/2}}% {3}+C=\frac{2(x^{2}+3x+20)^{3/2}}{3}+C.$

Now another integral which is a little more difficult:

Example 0.5.

$\int\frac{\cos(\ln x)}{x}dx$

The inner function here is $u=\ln x$ and $du=\frac{1}{x}dx$ .

$\int\frac{\cos(\ln x)}{x}dx=\int\cos(u)\cdot\frac{1}{x}dx=\int\cos(u)du=\sin(u% )+C=\sin(\ln x)+C.$

Example 0.6.

$\int\frac{3x^{2}+14x+1}{x^{3}+7x^{2}+x+115}dx$

This function is also a typical example of integration with substitution. Whenever there is a fraction, and the numerator looks like the derivative of the denominator, we set $u$ to be the denominator:

$u=x^{3}+7x^{2}+x+115,\quad du=(3x^{2}+14x+1)dx$

Thus:

$\int\frac{3x^{2}+14x+1}{x^{3}+7x^{2}+x+115}dx=\int\frac{1}{u}du=\ln u+C=\ln(x^% {3}+7x^{2}+x+115)+C.$

Example 0.7.

$\int\frac{7}{1+3x}dx$

As in the example above, we set $u=1+3x$ , $du=3dx$ :

$\int\frac{7}{1+3x}dx=\int\frac{7}{u}\frac{du}{3}=\frac{7}{3}\int\frac{1}{u}du=% \frac{7}{3}\ln u+C=\frac{7}{3}\ln(1+3x)+C.$

Example 0.8.

$\int t^{3}(t^{4}-50)^{700}dt$

Here the inner function is $u=t^{4}-50$ and $du=4t^{3}dt$ . Thus

$\int t^{3}(t^{4}-50)^{700}dt=\int\frac{u^{700}}{4}du=\frac{1}{4}\frac{u^{701}}% {701}+C=\frac{(t^{4}-50)^{701}}{4\cdot 701}+C.$

Some other examples (solve them!):

$\int e^{x}\sin(e^{x})dx,\quad\int\frac{e^{x}}{e^{x}+1}dx,\quad\int\frac{1}{x% \ln x}dx$

Title	a lecture on integration by substitution
Canonical name	ALectureOnIntegrationBySubstitution
Date of creation	2013-03-22 15:38:29
Last modified on	2013-03-22 15:38:29
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	4
Author	alozano (2414)
Entry type	Feature
Classification	msc 26A36
Related topic	ALectureOnIntegrationByParts
Related topic	ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution
Related topic	ALectureOnThePartialFractionDecompositionMethod