Substitution.

From theorem 5.9 about the derivative of the composition of two functions, we deduce the method of integration by substitution (i.e. change of variable).

Let $ f$ and $ g$ be two functions, each of them being differentiable on its own domain. By Theorem 5.9, $ gof$ is differentiable and we have $ (gof)'=g'of \cdot f'$ . It follows that:

$\displaystyle \int (gof)'(x) \; dx = \int (g'of)(x) \cdot f'(x) \; dx$ (8.1)

Denote $ t=f(x)$ ; then $ dt=f'(x) \; dx$ and Equation 5 can be written in the following form:

$\displaystyle \int (gof)'(x) \; dx = \int g'(f(x)) \cdot f'(x) \; dx = \int g'(t) dt$    

Example 8.2.4   Let $ F(x)= \int \frac {1}{4+x^2} \; dx$ .

Denote $ f(x)=\frac {1}{4+x^2}$ . We have:

$\displaystyle f(x)=\frac {1}{4(1+\frac {x^2}{4})}=\frac 14 \cdot \frac {1}{1+(\frac x2 )^2}.$    

Now let $ t=\frac x2$ ; then $ dt=\frac 12 dx$ , i.e. $ dx= 2 dt$ . We have:

$\displaystyle F(x)=\frac 14 \int \frac {1}{1+t^2} \; 2 dt = \frac 12 \int \frac {1}{1+t^2} \; dt = \frac 12 \arctan t +C =\frac 12 \arctan \frac x2 +C.$    

Example 8.2.5   Let $ F(x)= \int \frac {1}{1+x+x^2} \; dx$ .

We have: $ x^2+x+1= \left( x+ \frac 12 \right)^2+\frac 34
=\frac 34 \left[ \frac 43 \left( x+ \frac 12 \right)^2+ 1 \right]$ , thus:

$\displaystyle \frac {1}{1+x+x^2} = \frac {1}{\left( x+ \frac 12 \right)^2+\frac...
... 1 \right]} =\frac 43 \cdot \frac {1}{\frac 43 \left( x+ \frac 12 \right)^2+ 1}$    

Denote $ t=\frac {2 \sqrt{3}}{3}( x+ \frac 12)$ , thus $ dt=\frac {2 \sqrt{3}}{3} dx$ and we have:

$\displaystyle F(x)$ $\displaystyle =\frac 43 \int \frac {1}{t^2+1} \; \frac {\sqrt{3}}{2}dt$    
$\displaystyle \quad$ $\displaystyle = \frac {2 \sqrt{3}}{3} \int \frac {1}{t^2+1} \; dt$    
$\displaystyle \quad$ $\displaystyle = \frac {2 \sqrt{3}}{3} \arctan t +C$    
$\displaystyle \quad$ $\displaystyle =\frac {2 \sqrt{3}}{3} \arctan \left( \frac {\sqrt{3}}{3}(2x+ 1) \right) +C$    

Noah Dana-Picard 2007-12-28