Many functions have complicated derivatives, so trying to work backwards and find the antiderivative might prove impossible. The purpose of this section is to show one technique that helps work backwards from the Chain Rule. Recall the Chain Rule:
$$\frac{d}{dx}\left(f(g(x)\right)=f^\prime(g(x))\cdot g^\prime(x)$$
The goal is the determine: \(\displaystyle{\int f(g(x))g^\prime(x)dx}\).
By making a small change of variables, this original problem can be turned into a function which makes finding the antiderivative workable. Any variable can be used, typically the new variable is \(u\). By letting \(u=g(x)\), the INSIDE function, and taking the derivative of both sides of this substitution: \(\frac{du}{dx}=g^\prime(x)\implies du=g^\prime(x)dx\). Now the definite integral can be written as:
$$\int\left(f(\color{blue}{g(x)}\right)\color{green}{g^\prime(x)dx}\implies\int f(\color{blue}{u})\color{green}{du}$$
with the understanding that \(f\) will have an antiderivative easy to find.
| $$\int2(2x+1)^3dx$$ | $$\int10e^{10x}dx$$ | $$\int4x^3(x^4+5)^{10}dx$$ |
|---|---|---|
| \(\begin{array}{l} u=2x+1\implies du=2dx \\ \implies \int u^3du \\ \implies \frac{1}{4}u^4+C \\ \implies \frac{1}{4}(2x+1)^4+C\;\color{green}{\checkmark} \end{array}\) | \(\begin{array}{l} u=10x\implies du=10dx \\ \implies \int 10e^udu \\ \implies 10e^u+C \\ \implies 10e^{10x}+C\;\color{green}{\checkmark} \end{array}\) | \(\begin{array}{l} u=x^4+5\implies du=4x^3dx \\ \implies \int u^{10}du \\ \implies \frac{1}{11}u^{11}+C \\ \implies \frac{1}{11}(x^4+5)^{11}+C\;\color{green}{\checkmark} \end{array}\) |
The following video shows solutions worked out to the following problems:
Putting the last two topics together opens up a wide range of definite integrals. The following example will show substitution helping to evaluate a definite integral.
Evaluate the definite integral: \(\large{\int_{0}^{2}\frac{dx}{(x+1)^3}}\). There are two schools of thought on how to handle this type of problem. Below, both methods are discussed.
Ignore the limits of integration, use substitution and find antiderivative, change variables back and use FTC with original inegration limits: $$\begin{array}{l} \int\frac{dx}{(x+3)^2},\;u=x+3\implies du=1dx \\ \implies \int\frac{1}{u^3}du \\ \implies \frac{1}{-2}u^{-2}+C \\ \implies \frac{1}{-2}(x+3)^{-2}\bigg |_0^2 \\ \implies \frac{1}{-2}(2+3)^{-2}-\frac{1}{-2}(0+3)^{-2} \\ \implies \color{red}{\frac{1}{-2}(5)^{-2}-\frac{1}{-2}(3)^{-2}} \\ \implies \frac{1}{-2}\cdot\frac{1}{5^2}-\frac{1}{-2}\cdot\frac{1}{3^2} \\ \implies \frac{1}{-2}\cdot\frac{1}{25}-\frac{1}{-2}\cdot\frac{1}{9} \\ \implies \frac{-9}{450}+\frac{25}{450} \\ \implies \frac{16}{450}=\frac{8}{225}\;\color{green}{\checkmark} \end{array}$$
Use substitution, but also substitute the integration limits and use the variable \(u\) to finish. $$\begin{array}{l} \int_{0}^{2}\frac{dx}{(x+3)^2},\;u=x+3\implies du=1dx \\ x=0\implies u=3,\;x=2\implies u=5 \\ \implies \int_{3}^{5}\frac{1}{u^3}du \\ \implies \int u^{-3}du \\ \implies \frac{1}{-2}u^{-2}\bigg |_{3}^{5} \\ \implies \frac{1}{-2}(5)^{-2}-\frac{1}{-2}(3)^{-2} \end{array}$$ which is the same as the red row in METHOD ONE.