a lecture on integration by substitution

The Method of Substitution (or Change of Variables)

The following is a general method to find indefinite integrals that look like the result of a chain rule.

• When to use it: We use the method of substitution for indefinite integrals which look like the result of a chain rule. In particular, try to use this method when you see a composition of two functions.

• How to use it: In this method, we go from integrating with respect to $x$ to integrating with respect to a new variable, $u$, which makes the integral much easier.

  1. Find inside the integral the composition of two functions and set $u=$ “the inner function”.

  2. We also write $du=\frac{du}{dx}dx$.

  3. Substitute everything in the integral that depends on $x$ in terms of $u$.

  4. Integrate with respect to $u$.

  5. Once we have the result of integration in terms of $u$ ($+C$), substitute back in terms of $x$.

The method is best explained through examples:

Example 0.1.

We want to find $\int e^{2x}𝑑x$. The integrand is $e^{2x}$, which is a composition of two functions. The inner function is $2x$ so we set:

$$u=2x,du=2dx$$

Thus,

$$x=u/2,dx=du/2$$

Substitute into the integral:

$$\int e^{2x}𝑑x=\int e^u\frac{du}{2}=\frac{1}{2}\int e^u𝑑u=\frac{1}{2}{e}^u+C=\frac{1}{2}{e}^{2x}+C$$

The following are typical examples where we use the subsitution method:

Example 0.2.

$$\int x{e}^{3x^2+7}𝑑x$$

The inner function is $u=3x^2+7$ and $du=6xdx$. Thus $dx=du/(6x)$. Substitute:

$$\int x{e}^{3x^2+7}𝑑x=\int \frac{x{e}^u}{6x}𝑑u=\int \frac{e^u}{6}𝑑u=\frac{e^u}{6}+C=\frac{e^{3x^2+7}}{6}+C.$$

Example 0.3.

$$\int \sin (3x+7)𝑑x$$

The inner function is $u=3x+7$ and $du=3dx$. Therefore:

$$\int \sin (3x+7)𝑑x=\int \frac{\sin (u)}{3}𝑑u=-\frac{\cos (u)}{3}+C=-\frac{\cos (3x+7)}{3}+C.$$

Example 0.4.

$$\int (2x+3)\sqrt{x^2+3x+20}𝑑x$$

Inner $u=x^2+3x+20$ and $du=(2x+3)dx$. Thus:

$$\int (2x+3)\sqrt{x^2+3x+20}𝑑x=\int \sqrt{u}𝑑u=\int u^{1/2}𝑑u=\frac{2u^{3/2}}{3}+C=\frac{2{(x^2+3x+20)}^{3/2}}{3}+C.$$

Now another integral which is a little more difficult:

Example 0.5.

$$\int \frac{\cos (\ln x)}{x}𝑑x$$

The inner function here is $u=\ln x$ and $du=\frac{1}{x}dx$.

$$\int \frac{\cos (\ln x)}{x}𝑑x=\int \cos (u)\cdot \frac{1}{x}𝑑x=\int \cos (u)𝑑u=\sin (u)+C=\sin (\ln x)+C.$$

Example 0.6.

$$\int \frac{3x^2+14x+1}{x^3+7x^2+x+115}𝑑x$$

This function is also a typical example of integration with substitution. Whenever there is a fraction, and the numerator looks like the derivative of the denominator, we set $u$ to be the denominator:

$$u=x^3+7x^2+x+115,du=(3x^2+14x+1)dx$$

Thus:

$$\int \frac{3x^2+14x+1}{x^3+7x^2+x+115}𝑑x=\int \frac{1}{u}𝑑u=\ln u+C=\ln (x^3+7x^2+x+115)+C.$$

Example 0.7.

$$\int \frac{7}{1+3x}𝑑x$$

As in the example above, we set $u=1+3x$, $du=3dx$:

$$\int \frac{7}{1+3x}𝑑x=\int \frac{7}{u}\frac{du}{3}=\frac{7}{3}\int \frac{1}{u}𝑑u=\frac{7}{3}\ln u+C=\frac{7}{3}\ln (1+3x)+C.$$

Example 0.8.

$$\int t^3{(t^4-50)}^{700}𝑑t$$

Here the inner function is $u=t^4-50$ and $du=4t^{3}dt$. Thus

$$\int t^3{(t^4-50)}^{700}𝑑t=\int \frac{u^{700}}{4}𝑑u=\frac{1}{4}\frac{u^{701}}{701}+C=\frac{{(t^4-50)}^{701}}{4\cdot 701}+C.$$

Some other examples (solve them!):

$$\int e^x\sin (e^x)𝑑x,\int \frac{e^x}{e^x+1}𝑑x,\int \frac{1}{x\ln x}𝑑x$$

Title	a lecture on integration by substitution
Canonical name	ALectureOnIntegrationBySubstitution
Date of creation	2013-03-22 15:38:29
Last modified on	2013-03-22 15:38:29
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	4
Author	alozano (2414)
Entry type	Feature
Classification	msc 26A36
Related topic	ALectureOnIntegrationByParts
Related topic	ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution
Related topic	ALectureOnThePartialFractionDecompositionMethod