Subject: Calculus

Inverse Chain Rule Method

In differential calculus, we often mention chain rule in finding the derivative of a function especially composite functions. Now, we'll discuss inverse chain rule method. Inverse chain rule is a method of finding antiderivatives or integrals of a function by guessing the integral of that function, and then differentiating back using the chain rule. With that being said, let me introduce to you the following theorem.


Suppose that f is an integrable function and n is any number not equal to one. Then, the inverse chain rule is given by

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

So, if an integral is in the form reflected in left-side of the integral above, then we can easily find its derivative by doing inverse chain method. Also, most integrals like the one above can be also solved by doing substitution, that's why inverse chain rule anti-differentiation is also called as substitution method but a special case of substitution rule.


In this illustration, let us solve one integral of a function through inverse chain rule. Say, we are to integrate \int 3x^2(3+x^3)^8dx. Firstly, by looking at the integral, we can say that the integral, which is a product, can be integrated by simplifying the integrand into a polynomial. But you don't want to do that! Imagine a polynomial raised to eighth powers. That's a lot of terms.

Solution by inverse chain rule (theorem)

In this illustration, let us solve the integral of the above function through inverse chain rule. We can easily integrate such integral through inverse chain rule because it is in the form \int \left [f(x)\right ]^{n}f'(x)dx. Why? , the term 3+x^3 corresponds to f(x) and if we integrate 3+x^3 the answer is 3x^2dx which just corresponds to f'(x)dx regardless of the constant. Thus, according to the theorem above,

\int 3x^2(3+x^3)^8dx=\frac{(3+x^3)^{8+1}}{8+1}+C

Solution by substitution

By doing substitution, we let u=3+x^{3} such that du=3x^2dx. Inserting this into \int 3x^2(3+x^3)^8dx, we find that

\int x^2(3+x^3)^8dx=\int u^8du
\int u^8du = \frac{1}{9}u^9 + C

Inserting back u=3+x^{3}, we have

\frac{1}{9}u^9 + C=\frac{(3+x^3)^{9}}{9}+C

Which is just the same to our result doing inverse chain rule.

Example #1

Solve the following anti-derivative using inverse chain rule: \int \frac{\ln{x}}{x} dx

ANSWER: \int \frac{\ln{x}}{x} dx=\frac{1}{2}(\ln{x})^2+C

Example #2

Integrate using inverse chain rule method: \int 4x^3(\alpha_{o}+x^4)^c dx

Solution: \alpha_{o} is just a constant, so don't get intimidated by it. By the looks of it, this seems like an easy problem. Why? Because if you let f(x)=\alpha_{o}+x^4, then its derivative would be f'(x)=4x^3. Thus, our integration would be like

\int 4x^3(\alpha_{o}+x^4)^c dx = \int [f(x)]^cf'(x)dx

and by inverse chain rule method

\int \left [f(x)\right ]^{n}f'(x)dx = \frac{\left [f(x)\right ]^{n+1}}{n+1}+C.


\int 4x^3(\alpha_{o}+x^4)^c dx =\int [f(x)]^cf'(x)dx$}

= \frac{(\alpha_{o}+x^4)^{c+1}}{c+1}+C

ANSWER: \int 4x^3(\alpha_{o}+x^4)^c dx= \frac{(\alpha_{o}+x^4)^{c+1}}{c+1}+C

Example #3

Solve: \int \cos{x}\sin{x}dx

ANSWER: \int \cos{x}\sin{x}dx=\frac{1}{2}\sin^2{x}+C

Example #4

Calculate: \int e^{2\sin{x}}\cos{x}dx

Solution: First of all, we expand the integral into

\int e^{2\sin{x}}\cos{x}dx=\int e^{\sin{x}}\cdot e^{\sin{x}}\cos{x}dx

we then assign f(x)=e^{\sin{x}}, and its derivative f(x)=e^{\sin{x}}\cos{x}, the integral would then look like

\int e^{\sin{x}}\cdot e^{\sin{x}}\cos{x}dx = f(x)\cdot f'(x) dx .

Which is just the necessary condition for the inverse chain rule method formula. Thus,

\int e^{2\sin{x}}\cos{x}dx=\frac{1}{2}(e^{\sin{x}})^2+C

Example #5

Calculate: \int \frac{\cos{\ln{x}}}{x}dx

ANSWER: \int \frac{\cos{\ln{x}}}{x}dx=\sin{\ln{x}}+C

NEXT TOPIC: Substitution rule