Section F.2 A detailed strategy
Here is one detailed approach to evaluating integrals. It is based on the discussion in Section 3.5 and covers all the methods we have seen in class. For practice and test review, I strongly recommend selecting exercises from Section 3.5, so that you must decide what methods to use as well as then applying the methods correctly: that is the way integration is in real life.
Subsection F.2.1 Use tables of integrals and known integrals
The easiest approach is to recognize an integral as one you have already worked out how to evaluate. The corollary of this is that you should memorize the most common integrals, and collect the ones that you encounter often but have not memorized (yet) on a formula sheet. The formulas should be as flexible as possible with adjustable constants to avoid routine substitutions: for example, not \(\integral{\cos x}{x}=\sin x + C\) but instead
\begin{equation*}
\integral{\cos ax}{x}=\frac{1}{a}\sin ax + C
\end{equation*}
Subsection F.2.2 Do basic simplifications
Simplify first is a good strategy in many mathematical situations: try to simplify the function involved before starting on the calculus itself.
The most general basic simplifications are breaking up sums and differences into separate integrals, taking constant factors out in front of each integral (including division by constant factors), and rewriting roots as fractional powers.
It is also often useful to eliminate divisions by rewriting powers in the denominator as negative powers in the numerator, and using trig identities like converting a factor \(\cos x\) in the denominator into a factor \(\sec x\) in the numerator.
For example, \(\dsintegral{\frac{x^2}{7} - \frac{1}{\sin 2x} + \frac{5x}{\sqrt[3]{1+x^2}}}{x}\) could be rewritten as
\begin{equation*}
\frac{1}{7}\integral{x^2}{x} - \integral{\csc 2x}{x} + 5\integral{x(1+x^2)^{-1/3}}{x}
\end{equation*}
One important special type of simplification is used with integrals of products of powers of trigonometric functions, which will be discussed below.
Subsection F.2.3 Substitution
If the above steps do not give the solution, the easiest of the two most powerful general tools is substitution, especially with some compositions and products. That is, finding a function \(u(x)\) so that the integrand is in the form \(f(u)u'(x)\text{,}\) or \(f(u)\ds\frac{du}{dx}\text{.}\) Then you can use the ``cancellation of differentials'' idea
\begin{equation*}
\integral{f(u) \frac{du}{dx}}{x} = \integral {f(u)}{u}
\end{equation*}
to get a new, simpler integration problem.
That puts you back at the beginning with a new hopefully simpler integral. If the new integral is not easier to evaluate, the substitution was not useful, so try something else: another substitution or a different method.
Subsection F.2.4 Choosing a substitution function \(u(x)\)
As \(u\) will often appear inside a composition, one common choice is the function inside a power or other composition. In the example above of \(5x(1+x^2)^{-1/3}\text{,}\) you could try \(u=1+x^2\text{.}\)
You must then check if the rest of the term is the derivative of \(u\text{,}\) up to a constant factor. In the current example \(du/dx = 2x\) and the remainder of the term apart from the power of \(u\) is \(5x\text{,}\) which does match the derivative up to a harmless constant factor of \(5/2\text{,}\) so this substitution will work. (Exercise: complete this integral.)
In general, you must look to see if the term to be integrated consists of just the derivative of \(u\) times some expression that can be put in terms of \(u\) only (with no stray \(x\) terms.) For \(\integral{\sin(x) \cos(x)}{x}\text{,}\) the substitution \(u=\sin(x)\) gives
\begin{equation*}
\dsintegral{u \frac{du}{dx}}{x} = \dsintegral{u}{u} = \frac{u^2}{2} + C = \frac{1}{2}\sin^2(x) + C
\end{equation*}
and also substitution \(u=\cos(x)\) gives
\begin{equation*}
\dsintegral{-u \frac{du}{dx}}{x} = -\dsintegral{u}{u} = -\frac{u^2}{2} + C = -\frac{1}{2}\cos^2(x) + C
\end{equation*}
which despite initial appearances (“\(u^2/2\)” vs “\(-u^2/2\)”) agree, due to \(\cos^2(x) + \sin^2(x) = 1\text{.}\)
This shows that there can be more than one useful substitution.
Subsection F.2.5 Integration by Parts
Our second and last general tool is Integration by Parts, as summarized by
\begin{equation*}
\int{u}{\,dv} = uv - \int{v}{\, du}.
\end{equation*}
Here \(u\) and \(v\) are both functions of the actual integration variable \(x\text{,}\) so that in more detail the rule is
\begin{equation*}
\int{u(x) \frac{dv}{dx}}{\, dx} = u(x)v(x) - \int{v(x) \frac{du}{dx}}{\,dx}
\end{equation*}
Note that if one does a definite integral directly, the formula becomes
\begin{equation*}
\int_{x=a}^{x=b}{u}{\,dv} = \Big[uv\Big]_{x=a}^{x=b} - \int_{x=a}^{x=b}{v}{\,du}
\end{equation*}
The word “parts” refers to the fact that only one part of the integrand gets integrated, at least initially: \(dv/dx\) is integrated to find \(v\text{,}\) while \(u\) gets differentiated. The main hint that this method might be useful is that the function is a product of several functions and we know how to integrate at least one of them: particularly, powers of \(x\text{,}\) trigonometric functions, exponentials, logarithms, and inverses of familiar functions. There are always many different ways to choose which factor is to be the factor \(u\) to be differentiated (with all the rest of the integrand going into the factor \(dv/dx\) to be integrated; or in other words into the differential \(dv\)). Some guidelines are
- Try to integrate the most complicated part you can.
- It it necessary that you can integrate the function that goes into the differential \(dv\text{.}\)
- It is desirable for the function \(u\) to have a simple derivative, and three very common choices are positive integer powers of \(x\text{,}\) logarithms, and any inverses of familiar functions.
Subsection F.2.6 Inverse Substitution, especially with trigonometric functions
All substitutions can be done in inverse form, where one specifies \(x=g(u)\text{,}\) \(dx=g'(u)du\) instead of \(u=f(x)\text{,}\) \(du = f'(x)dx\text{:}\) \(g\) is the inverse of \(f\text{.}\) This has the great advantage that the formulas for \(x\) and \(dx\) automatically put everything in terms of the new variable \(u\text{:}\) there are never any stray \(x\) terms. Of course, the inverse \(g\) might be a messier function to work with, so this method is at its best when \(f\) is itself the inverse of a familiar function, and the most common examples are when the forward substitution function \(f\) is an inverse trigonometric function, so that \(g\) is a basic trigonometric function: therefore, the new variable is traditionally called \(\theta\) instead of \(u\text{.}\) The three main cases are
- For an integral containing integer powers of \(\sqrt{a^2-x^2}\) and of \(x\text{,}\) use \(x = a \sin \theta\text{,}\) so \(dx = a \cos \theta \, d\theta\) and \(\sqrt{a^2-x^2} = a \cos \theta\text{.}\)
- For an integral containing integer powers of \(\sqrt{a^2+x^2}\) and of \(x\text{,}\) use\\ \(x = a \tan \theta\text{,}\) so \(dx = a \sec^2 \theta \, d\theta\) and \(\sqrt{a^2+x^2} = a \sec \theta\text{.}\)
- For an integral containing integer powers of \(\sqrt{x^2-a^2}\) and of \(x\text{,}\) use\\ \(x = a \sec \theta\text{,}\) so \(dx = a \sec \theta \tan \theta \, d\theta\) and \(\sqrt{x^2-a^2} = a \tan \theta\text{.}\)
In each case, an appropriate right-triangle summarizes all the formulas needed. The resulting integrals involve products of powers of trigonometric functions, and often the methods of section~F below are needed to evaluate them. In that case you can get solutions involving sine and coinse on multiples of a new variable \(\theta\text{.}\) These can be put in erms of just \(\sin \theta\) and \(\cos \theta\) using the double angle formulas
\begin{equation*}
\sin 2 \theta = 2 \sin \theta \cos \theta, \qquad \cos 2 \theta = 2 \cos^2 \theta - 1, \quad = 1 - 2\sin^2 \theta
\end{equation*}
Subsection F.2.7 Special simplifications and substitutions for products of trigonometric functions
It often helps to convert into an expression with just sines and cosines using \(\tan x = \sin x/\cos x\) and so on. The example above of \(\int{\sin x \cos x}{dx}\) illustrates some important special trigonometric substitutions: With products of integer powers of sine and cosine
- if there is an odd power of cosine one can gather a term like \(\cos x dx\) at the end and use the substitution \(\sin x = u\text{,}\) \(\cos x \, dx = du\text{,}\) leaving over an even power of \(\cos x\) which can be written as an integer power of \(\cos^2 x\) which in turn gets converted with\begin{equation*} \cos^2 x = 1 - \sin^2 x = 1 - u^2 \end{equation*}Then all the remaining \(x\) terms are in terms of \(\sin x\text{,}\) which becomes \(u\text{,}\) so you get an integral of a polynomial (easy), or rational function if there were some negative powers (see below).
- if there is an odd power of sine, one can gather a term like \(\sin x dx\) and use the substitution {\boldmath \(\cos x = u\text{,}\) \(\sin x\, dx = -du\)}, and deal with the remaining even power of \(\sin x\) using\begin{equation*} \sin^2 x = 1 - \cos^2 x = 1 - u^2. \end{equation*}
- if there are products of even powers of both sine and cosine (including the case where only one of these functions is present, like \(\sin^4 x\)), one can reduce the powers using the half-angle formulas\begin{equation*} \cos^2 x = \frac{1}{2}(1 + \cos 2x), \quad \sin^2 x = \frac{1}{2}(1 - \cos 2x) \end{equation*}Then expand, simplify, and if necessary apply further trigonometric simplifications to some parts. Eventually you will get an odd power of sine or cosine in each term, plus a constant. For example\begin{equation*} \begin{split} \sin^2 x \cos^2 x \amp= \frac{1}{2}(1 - \cos 2x)\frac{1}{2}(1 + \cos 2x) \\ \amp= \frac{1}{4}(1 - \cos^2 2x) \\ \amp= \frac{1}{4}\sin^2 2x \\ \amp= \frac{1}{4} \cdot \frac{1}{2}(1 - \cos 4x) \\ \amp= \frac{1}{8} - \frac{1}{8}\cos 4x \end{split} \end{equation*}
Subsection F.2.8 Integration of rational functions (ratios of polynomials)
It is possible to integrate any rational function \(\ds f(x) = \frac{P(x)}{Q(x)}\) if you can factorize the polynomial \(Q(x)\) in the denominator into linear factors \((x-r)\) from roots plus irreducible quadratic factors \(x^2 + bx +c\text{.}\) Irreducible means that the quadratic has no real roots, which by the discriminant test means that \(b^2 < 4c\text{.}\)
It is easy and convenient to divide \(P(x)\) and \(Q(x)\) by a constant so that the lead coefficient in \(Q(x)\) is one; this form is assumed from now on.
Also, if the numerator has the same or higher degree as the denominator, the function should first be simplified to the sum of a polynomial plus a proper rational function, one with numerator of lower degree than the denominator. This is done by synthetic division of polynomials.
The integration is done by rewriting the rational function as the sum of a polynomial plus terms that can be integrated with a few basic rules. The main integration rules needed are
\begin{equation*}
\int \frac{du}{u-r} = \ln |u-r| + C
\end{equation*}
\begin{equation*}
\int \frac{du}{(u-r)^n} = \frac{-1}{n-1}\frac{1}{(u-r)^{n-1}}+ C, \quad n > 1
\end{equation*}
\begin{equation*}
\int \frac{du}{a^2 + u^2} = \frac{1}{a} \arctan \frac{u}{a} + C
\end{equation*}
\begin{equation*}
\int \frac{u}{a^2 + u^2} du = \frac{1}{2} \ln (a^2 + u^2) + C
\end{equation*}
plus occasionally completing the square to get into the latter two forms.
- The first step is to convert to the case of a proper rational function, one with the polynomial \(P(x)\) in the numerator of degree less than \(Q(x)\text{,}\) if this is not already true.The polynomial part is easy to integrate, so I will deal from now on only with proper rational functions.\begin{equation*} \frac{x^3 + 3x + 1}{x^2-4} = \frac{x(x^2-4) + 7x + 1}{x^2 - 4} = x + \frac{7x+1}{x^2-4}. \end{equation*}
-
Next, factorize the denominator into linear and irreducible quadratic factors.
Example F.2.2.
The denominator in Example 1 has roots 2 and -2, and so\begin{equation*} \frac{7x+1}{x^2-4} = \frac{7x+1}{(x-2)(x+2)}. \end{equation*}Sometimes the factorization will have to involve quadratic factors that have no real roots and so cannot be written in terms of two linear factors. This is true if the discriminant \(b^2-4c < 0\text{.}\)For an irreducible quadratic factor \(x^2+bx+c\text{,}\) complete the square to get the form \(x^2+bx+c = (x+e)^2 + a^2\text{,}\) where \(e=b/2\text{,}\) \(a^2=c-b^2/4\text{.}\)Example F.2.3. \(\ds\frac{13}{x^3-4x^2+13x}\).
The denominator here has only one real root \(0\text{:}\)\begin{equation*} \frac{13}{x^3-4x^2+13x} = \frac{13}{x(x^2-4x+13)} = \frac{13}{x[(x-2)^2+3^2]}. \end{equation*} - Next the rational function can be written as a sum of constant multiples of simple functions, for which we know the integrals. The simple functions needed are as follows
- For a factor \((x-r)\) in the denominator, a term like\begin{equation*} \frac{A}{x-r} \end{equation*}
- For a factor \((x-r)^n\) in the denominator, \(n\) terms\begin{equation*} \frac{A_1}{x-r}, \quad \frac{A_2}{(x-r)^2}, \mbox{ and up to } \frac{A_n}{(x-r)^n} \mbox{ (the power in the original denominator)} \end{equation*}
- For an irreducible quadratic factor \((x+e)^2 + a^2\) in the denominator, two terms, which can be put together as\begin{equation*} \frac{A+Bx}{a^2+(x+b)^2} \end{equation*}
- For the most complicated case of a repeated irreducible quadratic factor like \([a^2+(x+b)^2]^n\text{,}\) put together the previous two ideas: a succession of terms each with a linear factor like \(A+Bx\) on top, and powers of \(a^2+(x+b)^2\) on the bottom, ranging up to the same \(n\)-th power as in the original denominator.
Example F.2.4.
For Examples 2 and 3,\begin{equation*} \frac{7x+1}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}. \end{equation*}and\begin{equation*} \frac{13}{x[(x-2)^2+3^2]} = \frac{A}{x} + \frac{B+Cx}{(x-2)^2+3^2} \end{equation*} - The constants in the expansion can be determined, and the first step is to clear the denominator: multiply through by the denominator \(Q(x)\) of the original rational function to get an equation between two polynomials.
Example F.2.5.
For the two examples above we get\begin{equation*} 7x+1 = A(x+2) + B(x-2) \end{equation*}and\begin{equation*} 13 = A[(x-2)^2+3^2] + (B+Cx)x. \end{equation*} -
To find the numerical values of the constants in this equation, one method (not my favorite) is to expand out the right hand side into multiples of powers of \(x\text{,}\) including a constant, set the coefficients of each power of \(x\) equal to those at left, and solve the resulting simultaneous equations for the constants \(A\text{,}\) \(B\text{,}\) etc.However this equation solving can be made easier or avoided entirely by the strategy of strategic substitution: first looking at the easier equations you get by substituting each root of the denominator \(Q(x)\) in for \(x\text{.}\)When there are repeated factors or irreducible quadratic factors, this substitution method will only give some of the constants. To get the others, one can seek other easy equations by substituting in a few other “nice” integer values such as \(x=0,1,-1, \dots\text{.}\)
Example F.2.6.
For the first example function above, \(x=-2\) and \(x=2\) give\begin{equation*} -13 = B(-4), \quad 15 = A(4) \end{equation*}which immediately give \(B=13/4\text{,}\) \(A=15/4\text{:}\)\begin{equation*} \frac{7x+1}{(x^2-4)} = \frac{7x+1}{(x-2)(x+2)} = \frac{15/4}{x-2} + \frac{13/4}{x+2}. \end{equation*}Example F.2.7.
For the second example, \(x=0\) is the only root to substitute in, giving \(13 = A[4+9]\) so \(A=1\text{.}\) Using this \(A\) value and substituting also \(x=1\) and \(x=-1\) gives\begin{equation*} 13 = [1+9] + (B + C), 13 = [9+9] - (B-C) \end{equation*}or \(B + C = 3\text{,}\) \(B - C = 5\text{,}\) with the solution \(B = 4\text{,}\) \(C= -1\text{:}\)\begin{equation*} \frac{13}{x^3-4x^2+13x} = \frac{13}{x[(x-2)^2+3^2]} = \frac{1}{x} + \frac{4-x}{(x-2)^2+3^2}. \end{equation*}The integral of this can be found using the four basic forms above plus a substitution \(u=x-2\text{.}\) (You get \(\ds\ln|x| + \arctan{((x-2)^2+3^2)} -\frac{1}{2}\ln((x-2)^2+3^2) + C\text{.}\)) - Finally, you can evaluate the resulting integrals, using the four basic integrals above:\begin{equation*} \int{\frac{7x+1}{(x^2-4)}}{\, dx} = \int{\frac{15/4}{x-2} + \frac{13/4}{x+2}}{\, dx} = \frac{15}{4}\ln|x-2| + \frac{13}{4}\ln|x+2| + C \end{equation*}and\begin{equation*} \int{\frac{13}{x^3-4x^2+13x}}{\, dx} = \int{\frac{1}{\, dx} + \frac{4-x}{(x-2)^2+3^2}}{\, dx} = \ln|x| + \arctan{((x-2)^2+3^2)} -\frac{1}{2}\ln((x-2)^2+3^2) + C \end{equation*}where the substitution \(u=x-2\) is needed for the last term.
Subsection F.2.9 Final steps: make sure that you answer the original question
Remember a few things that must be done before you are finished:
- Substitute if necessary to get the solution in terms of the original variable.
- For an indefinite integral, add the constant of integration: if \(F(x)\) is an antiderivative of \(f(x)\text{,}\)\begin{equation*} \int{f(x)}{\, dx} = F(x) + C. \end{equation*}
- For a definite integral, evaluate at the limits of integration:\begin{equation*} \int_a^b {f(x)} {\, dx} = [F(x)]_a^b = F(b) - F(a). \end{equation*}
- For a definite integral using substitution \(u=g(x)\text{,}\) you might instead do the evaluation in the new \(u\) variable to avoid converting back to the original variable:\begin{equation*} \int_{x=a}^{x=b} f(g(x)) g'(x) \, dx = \int_{u=c}^{u=d} f(u) \, du, \; c=g(a), \, d=g(b). \end{equation*}
- Similarly, for a definite integral using inverse substitution \(x=h(t)\text{,}\) you might do the evaluation in the new variable using.\begin{equation*} \int_{x=a}^{x=b} f(x) \, dx = \int_{t=c}^{t=d} f(h(t)) h'(t) \, dt, \end{equation*}But you have to solve equations to get the new limits, \(c\) from \(a\) and \(d\) from \(b\text{:}\) \(h(c)=a\text{,}\) \(h(d)=b\text{.}\)
