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The Substitution Method: Chain Rule in Reverse
Recall that we were motivated by the fundamental theorem of calculus to define the indefinite integral
Zf(x)dx
as all antiderivatives of f(x). That is, if F(x)is an antiderivative of f(x)then we define
Zf(x)dx =F(x) + C
for some constant C.
As a simple example we can compute the following indefinite integral.
Zcos xdx =sin x+C
This is easy because we know that F(x) = sin xis an antiderivative of f(x) = cos x.
So knowing the derivative of sin xlet me integrate cos x. So what about doing some more interesting
derivtives. Consider for instance
F(x) = sin(x3+2x)
Then using the chain rule we see that
F0(x) = cos(x3+2x)(3x2+2)
Well, this means that we now know how to integrate the function f(x) = cos(x3+2x)(3x2+2). Indeed we
get
Zcos(x3+2x)(3x2+2)dx =sin(x3+2x) + C
This integrand is more complicated than just cos x, but really the way we solved it was not much harder.
We just used the chain rule!
Some theory
Okay - let’s try to generalise this example. Suppose that our antiderivative is a composition of two functions.
F(x) = f(g(x))
Then we differentiate using the chain rule to get
F0(x) = f0(g(x))g0(x)
This tells us, by our definition of the indefinite integral, that
ZF0(x)dx =F(x) + C(by definition)
i.e., Zf0(g(x))g0(x)dx =f(g(x)) + C
This is a lot of symbols, so we make our life easier and write u=g(x). Plugging this in gives us
Zf0(u)du
dx dx =f(u) + C
But we also have that Zf0(u)du =f(u) + C
