# a lecture on integration by substitution

## The Method of Substitution (or Change of Variables)

• 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 .

• 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 ALectureOnIntegrationBySubstitution 2013-03-22 15:38:29 2013-03-22 15:38:29 alozano (2414) alozano (2414) 4 alozano (2414) Feature msc 26A36 ALectureOnIntegrationByParts ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution ALectureOnThePartialFractionDecompositionMethod