Integration by substitution
From Maths
TODO: Flesh this page out with explanation and Jacobian reference
1 dimensional example
Question: [math]\int_0^3\frac{4x}{(1+x^2)^2}dx[/math]
That [ilmath]1+x^2[/ilmath] is the problem, and the derivative is clearly something to do with just [ilmath]x[/ilmath] (infact [ilmath]=2x[/ilmath]) so this invites a substitution.
- Substitute [math]u=1+x^2[/math] then [math]\frac{du}{dx}=2x[/math] thus [math]du=2x dx[/math]
- Notice our integral is [math]I=\int_0^3\frac{4x}{(1+x^2)^2}dx=\int_0^3\frac{2}{(1+x^2)^2}2xdx[/math] we can literally replace the [ilmath]2xdx[/ilmath] with [ilmath]du[/ilmath] - this is a nice shortcut.
- More formally we actually have [ilmath]dx=\frac{du}{2x}[/ilmath] let us substitute this
- [math]\int_0^3\frac{4x}{(1+x^2)^2}dx=\int^{x=3}_{x=0}\frac{4x}{u^2}\frac{du}{2x}[/math]
- Notice: I write explicitly the limits of the integral because I have yet to express them in terms of [ilmath]u[/ilmath]
- [math]=\int^{x=3}_{x=0}\frac{2}{u^2}du=2\int^{x=3}_{x=0}\frac{1}{u^2}du[/math]
- Converting the limits:
- [ilmath]x=3\implies u=1+3^2=10[/ilmath]
- [ilmath]x=0\implies u=1[/ilmath]
- Converting the limits:
- Thus [math]I=2\int^{10}_{1}\frac{1}{u^2}du[/math] which the reader ought to be able to integrate.
- [math]\int_0^3\frac{4x}{(1+x^2)^2}dx=\int^{x=3}_{x=0}\frac{4x}{u^2}\frac{du}{2x}[/math]