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User:Graemeb1967/Introduction to Calculus Introduction to Integration

Integration by Substitution

= Introduction to this topic = This page is dedicated to teaching problem techniques specifically for integration by substitution. For other integration methods see other sources.

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

Theory of Integration by Substitution
This area is covered by the wikipedia article 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 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.

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

u = x-3\,\qquad \frac{du}{dx} = 1\,\qquad du = 1 dx\, $$ NOTE: that the differential of x-3 is 1.

Now we substitute u and du into the original integral.

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

Now 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\, $$

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\, $$

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
 * Superscript textSuperscript text$$

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

The Steps We Applied
Lets now review the steps for integration by substitution.
 * First identify that you have a function of a function. This skill comes with practise to identify candidates.
 * Identify u and then find du that is appropriate for the expression.
 * Integrate using normal techniques.
 * Substitute back the values for u.
 * Don't forget the constant of indegration for indefinate integrals.

Third examples
Consider the integral

\int_{0}^2 x \cos(x^2+1) \,dx $$ By using the substitution u = x2 + 1, we obtain du = 2x dx and



\begin{align} \int_{x=0}^{x=2} x \cos(x^2+1) \,dx & {} = \frac{1}{2} \int_{u=1}^{u=5}\cos(u)\,du \\ & {} = \frac{1}{2}(\sin(5)-\sin(1)). \end{align} $$

Here we substituted from right to left. It is important to note that since the lower limit x = 0 was replaced with u = 02 + 1 = 1, and the upper limit x = 2 replaced with u = 22 + 1 = 5, a transformation back into terms of x was unnecessary.