Stephen Ledbetter Blog Post 2/27/2011

This week went by pretty fast but the days seemed to just drag on and on and on. I'm going to explain how to work on integration by substitution.

If you have the integral of u * v or the integral of u/v, where v is the derivative of u, v will disappear.

Example:

the integral of (x^2 + 1)^2 (2x) dx

u = x^2 + 1

du = 2x dx

plug in to...

the integral of u^2 du

1/3u^3 + c

1/3(x^2 + 1)^3 + c

the integral of 5 cos 5x dx

u = 5x

du = 5 dx

plug in to...

the integral of cos u du

-sin u + c

-sin 5x + c

e^x

d/dx e^(x^3) = e^(x^3) * 3x^2

the integral of e^2x = e^2x/2 = 1/2 e^2x + c

the integral of x^2 e^(x^3) dx

u = x^3

du = 3x^2

1/3 * the integral of 3x^2 e^(x^3) dx

1/3 * the integral of e^u du

1/3 e^u/1 + c

1/3 e^(x^3) + c