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Integration by Substitution

[edit] Introduction to this topic

This page is dedicated to teaching problem solving techniques, specifically for integration by substitution. For other integration methods see other sources.

The format is aimed at first introducing the theory, the techniques, the steps and finally a series of examples which will make you further skilled.

[edit] Assumed Knowledge

Basic Differentiation
Basic Integration Methods
function of a function (w:composite function)
w:chain rule

[edit] Theory of Integration by Substitution

This area is covered by the wikipedia article w:Integration by Substitution. On this page we deal with the practical aspects.
We begin with the following as is described by the wikipedia article

$\int_a^b f(g(t))g'(t)\, dt = \int_{g(a)}^{g(b)} f(x)\,dx.$

This can be rewritten as

$\int_{}^{}f(u)\,du.$

by setting

$u = g(x)\,\qquad du = g'(x)\,dx.$

The principle applied here is function of a function (w:composite function) and the reverse of the w:chain rule, this is the basis of integration by substitution.
The key skill now is to identify what value we use for u and following the process to solution.

[edit] Technique

[edit] Example 1

Let us examine this integral

$\int_{}^{}(x-3)^{10}\,dx.$

The inner function is

$x-3\,$

The outer function is

$(\qquad)^{10}$

Recognising this relationship we then move onto the following set of steps to process the inner function
NOTE: that the differential of $x - 3$ is $1$ .

$u = x-3\,\qquad \frac{du}{dx} = 1\,\qquad du = 1 dx\,\qquad du = dx\,$

Now we substitute u and du into the original integral.

$\int_{}^{}(x-3)^{10}\,dx.\qquad \int_{}^{}(u)^{10}\,du.$

Then apply standard integral technique

$\frac{u^{n+1}}{n+1}+ c\,\qquad \frac{u^{11}}{11}+ c\,$

And finally we substitute the value of u back into the equation

$\frac{(x-3)^{11}}{11}+ c\,$

[edit] Example 2

Let us examine this integral

$\int \frac{x}{\sqrt{9+x^2}}\,dx.$

We can first rearrange the fraction to make it more familiar.

$\int x (9+x^2)^\frac{-1}{2}\,dx.$

The inner function is

$9+x^2\,$

The outer function is

$x (\qquad)^\frac{-1}{2}$

Next we assign u and du

$u = 9+x^2\,\qquad \frac{du}{dx} = 2x\,$

But we have a problem! $du\,$ doesnt equal $x\,$ ! So we need to rearrange our formula for $du\,$ .

$\frac{du}{dx} = 2x\,\qquad \frac{du}{2} = x dx\,\qquad \frac{1}{2}du = x dx\,$

Now we can substitute u and du into the original integral.

$\int x (9+x^2)^\frac{-1}{2}\,dx.\qquad \frac{1}{2}\int (u)^\frac{-1}{2}\,du.$

Study the above substitution carefully. We moved the fractional component of du to the front as it represents a constant.

Now apply standard integral technique

$\frac{u^{n+1}}{n+1}+ c\,\qquad \frac{1}{2}.\frac{u^\frac{1}{2}}{\frac{1}{2}}+ c\,$

Cleaning up this expression we have

$u^\frac{1}{2}+ c\,$

And finally we substitute the value of u back into the equation

$(9+x^2)^\frac{1}{2}+ c\,\qquad \sqrt{9+x^2}+ c\,$

[edit] The Steps We Applied

Lets now review the steps for integration by substitution.

	Indefinite Integral	Definite Integral
1.	First identify that you have a function of a function. This skill comes with practice to identify candidates.	First identify that you have a function of a function. This skill comes with practice to identify candidates.
2.	Identify u and then find du that is appropriate for the expression.	Identify u and then find du that is appropriate for the expression.
3.		Change limits for definite integrals.
4.	Integrate using normal techniques.	Integrate using normal techniques.
5.	Substitute back the values for u for indefinite integrals.
6.	Don't forget the constant of integration for indefinite integrals.

[edit] The Definite Integral

Consider the definite integral

$\int_{0}^2 x \cos(x^2+1) \,dx$

By using the substitution

$u = x^2+1 \qquad \frac {du}{dx} = 2x\, \qquad \frac {du}{2} = x\,dx$

Now because we have limits, we need to change them with respect to u. Note the value of the limits.

$\int_{x=0}^{x=2} x \cos(x^2+1) \,dx$

$\begin{align} & {} u=(0^2+1)=1 \\ & {} u=(2^2+1)=5 \end{align}$

Now we have a new definite integral to solve

$\begin{align} & {} = \frac{1}{2} \int_{u=1}^{u=5}\cos\,u \,du \\ & {} = \frac{1}{2}(\sin5-\sin1). \end{align}$

[edit] Finding u

Lets look at more examples at finding u.

[edit] Example 1

$\int_{0}^1 \frac {x \cosh\sqrt{4-x^2} } {\sqrt{4-x^2}} \,dx.$

$u\,$	$\frac {du}{dx}\,$	$Limits\,$	$u\,Integral\,$
$u =\sqrt{4-x^2}$	$\frac {du}{dx} = \frac {1}{2}(4-x^2)^\frac {-1}{2} \times -2x\,$	$-\int_{x=0}^{x=1} cosh\,u \,du.$	$-\int_{u=2}^{u=\sqrt{3}}\cosh\,u\,du$
	$= -x(4-x^2)^\frac {-1}{2}$	$u =\sqrt{4-1^2} = \sqrt{3}$	$= - (sinh\,2 - sinh\,\sqrt{3})$
	$= \frac {-x}{\sqrt{4-x^2}}$	$u =\sqrt{4-0^2} = 2$	$= - sinh\,2 + sinh\,\sqrt{3}$
	$du = \frac {-x}{\sqrt{4-x^2}}\,dx.$