I - Integration by Substitution Lesson

Integration by Substitution

Finding an antiderivative when familiar antidifferentiation/integral formulas do not apply is not always obvious. A more general technique for finding antiderivatives is needed. One such method is integration by substitution, frequently referred to as reversing the Chain Rule.

Chain of Variables

Recall that if u is a differentiable function of x and n is any number different from -1, then by the Chain Rule $\frac{d}{dx}=\left(\frac{u^{n+1}}{n+1}\right)=u^n\frac{du}{dx}$. Integrating both sides of this formula produces $\int\frac{d}{dx}\left(\frac{u^{n+1}}{n+1}\right)=\int u^n\frac{du}{dx}$, which in differential form is $\int d\left(\frac{u^{n+1}}{n+1}\right)=\int u^ndu$ or in the more familiar form $\int u^ndu=\frac{u^{n+1}}{n+1}+C$. This serves as a reminder that whenever an integral can be put in the form $\int u^n\frac{du}{dx}$ with u a differentiable function of x and du the differential of u, it is possible to use this general rule to integrate with respect to u. It was Isaac Newton, one of the founders of calculus, who first recognized the potential of rewriting difficult integrals as $\int u^n\frac{du}{dx}$ by changing the variable or variable expression to u and using a substitution as a viable approach to finding antiderivatives.

A change of variables requires rewriting an integral in terms of u and du (or any other convenient variable). If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then du = g'(x)dx, and $\int f\left(g\left(x\right)g'\left(x\right)\right)dx=\int f\left(u\right)du=F\left(u\right)+C$. Examples of integration using change of variables are presented below.

Example 1

$\int2x\cos\left(x^2\right)dx$

Solution: $Let \: u=x^2\\ Then \: \frac{du}{dx}=2x,du=2xdx\\ \int 2x\cos\left(x^2\right)dx=\int \cos udu=\sin u+C=\sin x^{2+c}$

Example 2

$\int\sqrt[]{3x-2}dx$

Solution: $Let\: y=3x-2\\ Then\: \frac{du}{dx}=3,du=3dx,dx=du\\ \int \sqrt[]{3x-2}dx=\frac{1}{3}\int u^{\frac{1}{2}}du=\frac{1}{3}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{9}u^{\frac{3}{2}}+C=\frac{2}{9}\left(3x-2\right)^{\frac{3}{2}}+C$

Example 3

$\int\sqrt[]{3x-2}dx$

Solution: $Let\:u=3x-2,\frac{\left(u+2\right)}{3}=x\\ Then\:\frac{du}{dx}=3,du=3dx,\frac{du}{3}=dx\\ x\sqrt[]{3x-2}dx\\ \int\frac{u+2}{3}\cdot u^{\frac{1}{^2}}\frac{du}{3}=\frac{1}{9}\int\left(u^{\frac{3}{2}}+2u^{\frac{1}{2}}\right)du\\ =\frac{1}{9}\left(\frac{u^{\frac{5}{2}}}{\frac{5}{2}}+\frac{2u^{\frac{3}{2}}}{^{\frac{3}{2}}}\right)+C=\frac{1}{9}\left(\frac{2u^{\frac{5}{2}}}{5}+\frac{4u^{\frac{3}{2}}}{3}\right)+C\\ =\frac{2u^{\frac{5}{2}}}{45}+\frac{4u^{\frac{3}{2}}}{27}+C=\frac{2\left(3x-2\right)^{\frac{5}{2}}}{45}+\frac{4\left(3x-2\right)^{\frac{3}{2}}}{27}+C$

Example 4

$\int\sec^2\left(3x-1\right)dx$

Solution:  $Let\:u\:=\:3x-1\\Then\:\frac{du}{dx}=3,\:du=3x,dx=\frac{du}{3}\\ \int\sec^2(3x-1)dx=\frac{1}{3}\int\sec^2udu=\frac{1}{3}\tan u+C=\frac{1}{3}\tan(3x-1)+C$

Finding integrals of more complicated trigonometric functions may also benefit from the substitution technique as illustrated in the three presentations below.

View the following presentation from the beginning to 6:24.

Change of Limits for Definite Integrals

When evaluating definite integrals, it is often more expedient to change the limits of integration at the beginning of the process rather than remembering to substitute the original variable back into the expression at the end of the process. If the function u = g(x) has a continuous derivative on the closed interval [a, b] and f is continuous on the range of g, then $\int_a^{b}f\left(g\left(x\right)\right)g'\left(x\right)dx=\int_{g(a)}^{g(b)}f\left(u\right)du$. The video below illustrates how definite integrals can be evaluated using the u-substitution technique involving change of limits.

Three additional examples of u-substitution involving change of limits definite integrals are presented below.

Example 1

$\int_0^{\frac{\pi}{4}}\sin\left(4t\right)dt$

Solution: Let u = 4t

Then:

$\frac{du}{dt}=4,du=4dt,dt=\frac{du}{4}\\ u_{upper}.\:When\:t\:=\frac{\:\pi}{4},\:u=\pi\\ u_{lower}.When\:t=0,\:u=0$

$\int_{0}^{\frac{\pi}{4}}\sin\left(4t\right)dt=\frac{1}{4}\int_0^\pi\sin udu=-\frac{1}{4}\cos u\Biggr|_0^\pi=\\ -\frac{1}{4}\left(\cos\Pi=\cos0\right)=-\frac{1}{4}\left(-1-1\right)=\frac{1}{2}$

Example 2

$\int_{-19}^8\sqrt[3]{8-x}dx$

Solution: Let u=8-x

Then: $\frac{du}{dx}=-1,\:du=-dx,\:dx=-du\\ u_{upper}.When\:x=8,\:u=0\\ u_{lower}.\:When\:x=-19,\:u=27$

$\int_{-19}^8\sqrt[3]{8-x}dx=-\int_{27}^0u^{\frac{1}{3}}du=\int_0^{27}u^{\frac{1}{3}}du=\frac{3}{4}u^{\frac{4}{3}}\Biggr|_0^{27}=\\ \frac{3}{4}\left(27^{\frac{4}{3}}-0\right)=\frac{243}{4}60.75$

Example 3

$\int_0^2\frac{x}{\left(1+x^2\right)^3}dx$

Solution: Let $y=1+x^2$

Then: $\frac{du}{dx}=2x,\:du=2xdx\\ u_{upper}.When\:x=2,\:u=5\\ u_{lower}.\:When\:x=0,\:u=1$

$\int_0^2\frac{x}{\left(1+x^2\right)^3}dx=\frac{1}{2}\int_1^5u^{-3}du=\frac{1}{2}\cdot\frac{u^{-2}}{-2}\Biggr|_1^5=\\ -\frac{1}{4}\cdot\frac{1}{u^2}\Biggr|_1^5=-\frac{1}{4}\left(\frac{1}{25}-1\right)=\frac{6}{25}=0.24$

Numerical Integration via Technology

Many graphers and mathematics software have built-in numerical integration tools that compute an approximation to the definite integral of f with respect to a given variable from a to b. For those who rely primarily on numerical integration tools, it is important to understand the limitations of the resident algorithms in order to judge how accurately the approximation differs from the exact answer.

The TI calculator family command for numerical integration is fnInt, and the required syntax is fnInt(f(variable), variable, a, b), where a is the lower limit of integration and b is the upper limit of integration.

Watch the following video to see how this function works.

Integration by Substitution Practice

Integration by Substitution: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of definite integrals and integration by substitution.

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