4.5a Integration by Substitution Use pattern recognition Use Change of Variables Use the General Power Rule for Integration Why? All to find an indefinite or definite integral!
We will focus on techniques to integrate composite functions. The main two techniques are pattern recognition and change of variables . Both require a u-substitution. The role of substitution in integration is much like the role of the chain rule in differentiation – in other words, it helps with more complex functions where there is an outside part and an inside part.
Let’s revisit the Chain Rule: For differentiable functions given by So going backwards with integration,
Now the trick comes in figuring out what is f(g(x)) and what is g(x)! You need to develop some experience to get good at it. Notice the integrand contains both f(g(x)) and g’(x)! p. 295
Let’s see if we can see some patterns! Each of these fits the pattern of f(g(x))g’(x). Identify f(g(x)) and g’(x)
These just need you to multiply and divide by a constant to do the same thing.
Notice that the composite function in the integrand has an outside function f and an inside function g, and a factor that would be g’ Ex 1 p.296 Recognizing the f(g(x))g’(x) pattern Find What is f(g(x))? What is g(x)? Do we have g’(x)? Do you recognize the general power rule for integration?
Ex 2 p. 296 Recognizing the f(g(x))g’(x) pattern What is f(g(x))? What is g(x)? Do we have g’(x)? cos 5x 5x 5 You can always check by differentiating!
So far in the examples the pattern fit exactly. In many integrals, the pattern is essentially there, you just need a constant to make it work. We would multiply and divide by a constant by using the constant multiple rule
Ex 3 p. 297 Multiplying and Dividing by a Constant This is almost the pattern, what are we missing? This idea ONLY applies to constants – we can’t multiply and divide by x, for example!
Sometimes things are complicated enough that we should just do a formal change of variables. where Just remember to change back at the end!
Ex 4 p298 Change of variables We don’t have a rule for the square root of a function. Let u=2x-1 du=2dx or dx = ? This is a problem because variables don’t agree! Change back to function of x at the end.
Ex 5 p. 298 Change of Variables Let u = x 3 – 1 du = 3x 2 dx What do I need that I don’t have?
Ex 6, p. 299 Change of variable Here the trick is to figure out what is the u and what is the u’ Let u = sin 4x then du = 4cos4x dx, which means I am missing a 4
I look for the most complicated part of the integrand and see if I have a u and du that would work to simplify it.