The Reverse Chain Rule

Expert reviewed 21 July 2024 5 minute read


HSC Maths Advanced Syllabus

  • establish and use the formula $\int f'(x)[f(x)]^ndx=\frac{1}{n+1}[f(x)]^{n+1}+Cwherewheren\ne 1$ (the reverse chain rule)

Note:

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What is the Reverse Chain Rule?

The Reverse Chain Rule is a method used to simplify the integration of composite functions. It is essentially the application of the chain rule of differentiation in reverse.

The Reverse Chain Rule involves identifying a part of the integrand that can be treated as the derivative of another function. This method is particularly useful when you're dealing with integrals of functions that are products of functions, one of which is the derivative of another function inside the integral. By recognising this relationship, you can simplify the integral significantly.

The formula for the reverse chain rule is as follows:

f(x)[f(x)]ndx=1n+1[f(x)]n+1+C\int f'(x)[f(x)]^ndx=\frac{1}{n+1}[f(x)]^{n+1}+C

How to use the Reverse Chain Rule

For the rule to be effective, we must manipulate a given function to match the formula. We can do this by using the previous rules for integration outlined earlier in this module.

Let’s walk through an example:

For the expression 9x2(x3+1)2dx\int9x^2(x^3+1)^2dx, we cannot yet apply the formula, as it is not in the correct form. To change it, we must first find f(x)f(x) and f(x)f'(x).

f(x)=x3+1f(x)=3x2f(x)=x^3+1\\f'(x)=3x^2

Now, looking at the expression, we can see that the term which should represent f(x)f'(x), which is 9x29x^2 does not match the value of 3x23x^2. However, we can adjust this by multiplying the expression by a value, to change 9x29x^2 to 3x23x^2. However, whatever operation we apply by multiplication to the expression inside the integral, we must do the opposite outside the integral.

313×9x2(x3+1)2dx=33x2(x3+1)2dx3\int \frac{1}{3}\times9x^2(x^3+1)^2dx=3\int3x^2(x^3+1)^2dx

Now that we have the expression in terms of the reverse chain rule formula, we can simply use the formula to solve the integral:

33x2(x3+1)2dx=3[12+1(x3+1)2+1+C=3[(x3+1)33]+C=(x3+1)3+C3\int3x^2(x^3+1)^2dx=3[\frac{1}{2+1}(x^3+1)^{2+1}+C\\=3[\frac{(x^3+1)^3}{3}]+C\\=(x^3+1)^3+C

Return to Module 4: Integration