Home  18.013A  Chapter 19 


By far the most important tool of antidifferentiation is that of applying the chain rule of differentiation backwards; that means given a function $f(x)$ , finding a function $u(x)$ , so that you can write $f(x)$ as $\frac{dg(u(x))}{du}\frac{du}{dx}$ . In that case you can claim that an antiderivative of $f$ is $g(u(x))$ .
(This seems complicated when written this way, but is easier to use than it looks, in practice.)
For example, suppose we have $f(x)={\mathrm{sin}}^{5}x\mathrm{cos}x$ .
Then we can set $u(x)=\mathrm{sin}x$ , and find that $f(x)={u}^{5}\frac{du}{dx}$ , which we can recognize as the derivative of $\frac{{u}^{6}}{6}$ or $\frac{{\mathrm{sin}}^{6}x}{6}$ , which is then an antiderivative of $f$ .
The identical idea can be used to antidifferentiate any polynomial in sines and cosines of odd degree in all, if you apply the identity ( ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$ ) to bring all terms into forms that are linear in one of sine or cosine like $f$ is above.
Employing this technique is called "substituting" ; the variable $u$ is in a sense substituted for $x$ .
Notice that if we can recognize that $f(x)$ is the derivative of a known function then we can antidifferentiate by inspection.
Using substitution of $u(x)$ for $x$ , we change the question to: can we recognize $\frac{f(x(u))}{du/dx}$ as the derivative of a function of $u$ ?
In the example above, the new question becomes: can we recognize ${\mathrm{sin}}^{5}x$ as a derivative with respect to the variable $\mathrm{sin}x$ ? Since ${\mathrm{sin}}^{5}x$ is a power of $\mathrm{sin}x$ , we can recognize it as the derivative of $\frac{{\mathrm{sin}}^{6}x}{6}$ and we have found an antiderivative of $f$ .
There are some well known standard substitutions that allow us to antidifferentiate many potentially useful functions. We review them below.
Looking for the right substitution that might help for finding some given specific antiderivative is a bit of intellectual detective work. It is akin to solving a chess problem, in that while perhaps it is of no practical value, it may well be an excellent exercise for developing your reasoning powers or your ability to find what at first seems like finding a needle in a haystack.
Here are classes of functions (called integrands) that you should be able to integrate at least in theory. The details can become ugly and boring but that merely means that you cannot be asked to do them out very often.
1. Any polynomial in $x$ . These can be done directly term by term by inspection using the power rule backwards.
2. Any single power positive or negative in $x$ , such as ${(x3)}^{1/3}$ or ${(1x)}^{1}$ . Same method.
3. Any exponent like ${e}^{ax}$ . Again direct integration can be applied using the rule for differentiating exponents backwards.
4. Any polynomial in sines and cosines multiplied by any exponent as in the last case. Here you can use the exponential formula for sines and cosines to make this into an ugly sum of exponentials to each of which case 3 applies. Some of the exponents will be complex, but so what.
5. The function $\frac{1}{1+{(xa)}^{2}}$ , whose integral is the arctangent of $(xa)$ .
6. Any rational function of $x$ which means any function of the form $\frac{p(x)}{q(x)}$ where $p$ and $q$ are both polynomials. To do this you must be able to factor $q$ into linear (and perhaps quadratic terms), employ the technique called "partial fraction expansion" discussed in a later section, and then integrate using cases 1 and 2 and perhaps 5 above.
(The idea is that $\frac{p}{q}$ can be written as a sum of a polynomial and a sum over the zeroes of $q(x)$ of single inverse powers (and perhaps terms like that in case 5 if you don't like to use complex zeroes).)
7. Any rational function of sines and cosines of $x$ (in theory anyway). This reduces to a rational function to which case 6 applies after the magic substitution $t=\mathrm{tan}\left(\frac{x}{2}\right)$ . The thought of carrying this to completion for a complicated function is too horrible for me to contemplate, but it ought to work. Theoretically the same thing should work for rational functions of sinh's and cosh's using the substitution $t=\mathrm{tanh}\left(\frac{x}{2}\right)$ .
8. A rational function of any of $\sqrt{1{x}^{2}}$ and $x$ , or of $\sqrt{{x}^{2}1}$ and $x$ , or of $\sqrt{{x}^{2}+1}$ and $x$ . To do these you use the substitution $x=\mathrm{sin}t$ , or $x=\mathrm{cosh}t$ or $x=\mathrm{sinh}t$ , respectively to obtain a rational function of sines and cosines of $t$ in the first case and of sinh's and cosh's in the latter two cases. You can then apply case 7 to finish the job.
9. A rational function of $x$ and of one square root, of either a linear or quadratic function. (An ugly example is $\frac{1}{{\left(x+\sqrt{{x}^{2}+ax+b}\right)}^{2}}$ ; here is another $\frac{1}{\sqrt{x3}{x}^{2}}$ .)
If you have a square root of a linear factor you can use the substitution $t=$ that square root factor, to convert your integrand to a rational function, which you can then integrate using case 6.
With a quadratic factor you complete the square and then make linear substitutions to get your integrand into one of the forms of case 8.
It is a good idea to be able to recognize the classes of integrands to which these methods might be applied. However, unless you never make errors in doing anything, I suggest you use Maple or Mathematica or something like that to actually do complicated integrals of the later kinds if you are forced to do one.
There are some easy examples you could practice on to get an idea what is involved.
10. The product of a sine or cosine or exponent in $x$ and a polynomial in $x$ .
This reduces to doing the product of sine, cosine or exponent and a single power for each power in the polynomial and adding the results.
The latter integrals can be done either by integrating by parts or differentiating with respect to a parameter, both of which are discussed below.
11. The product of an exponent, a polynomial in sines and cosines and a polynomial in $x$ . This can be reduced to case 10 by using the expression for trigonometric functions as complex exponentials.
12. Integrals involving logarithms and other functions. You can get rid of the logarithms by integrating by parts, and then, if you get something you can recognize above, you can finish the job.
Are these all the functions that we can integrate?
No, there are others. (You can generate arbitrarily complicated functions that you can integrate by writing down horrible looking functions and differentiating them. You will find it easy to integrate the results.)
And every integrand consisting of standard functions that are not singular where being integrated can be integrated numerically to whatever accuracy you want. The advantage of the methods described above over numerical integration is that you can keep parameters indefinite here, and must specify them when doing a straightforward numerical integration.
In addition there is a large class of special functions whose properties have been extensively studied, and a large number of integrals that can be expressed in terms of them.
In this modern age it is a good idea to cultivate the art of recognizing which integrands look integrable to you. If the function you want to integrate does not look familiar to you try one of the mathematics programs.
