Strategy for integration

Strategy for integration

  • Up to now, we have been learning methods of integration one by one:
    • substitution
    • trigonometric integration
    • partial fractions
  • Now, we'll practice figuring out how to calculate an integral without knowing beforehand what method you will have to use.
    • As I said before, integration is not an algorithmic process.
    • Moreover, the same integral can be calculated via different methods.
    • But there's a few steps you can take which can help you figure out how to calculate an integral.
  • As a zeroth step, you absolutely have to know by heart the basic integration formulas. For example, you can find a table in the textbook, on the first page of section 7.5.

I. Simplify the integrand if possible.

  • The integrand might be of a different form then one to which one of the methods can be readily applied.
    • That is, if you don't immediately know what to do with the integrand, you can try algebraic manipulations.
    • Example. 7.5.4. Since we have \[ \frac{\sin^3x}{\cos x}=\frac{\sin x(1-\cos^2x)}{\cos x}=\tan x-\sin x\cos x=\tan x-\frac12\sin2x, \] we get \[ \int\frac{\sin^3x}{\cos x}\,\mathrm dx=\int\tan x-\frac12\sin2x\,\mathrm dx=-\ln|\cos x|+\frac14\cos2x+C. \]

II. Look for an obvious substitution.

  • You should always be on the lookout for whether the integrand can be recognized as the product of the form \(kf(g(x))g'(x)\).
    • Example. 7.5.32. In the integral \(\int_1^3\frac{e^{3/x}}{x^2}\,\mathrm dx\), we see a product of the composite function \(\exp(\frac{3}{x})\) and the function \(x^{-2}\). Therefore, we can interpret the integrand as \(kf(g(x))g'(x)\) where \(f(x)=e^x\), \(g(x)=\frac{3}{x}\), and \(k=-\frac16\). Therefore, we have \[ \int_1^3\frac{e^{3/x}}{x^2}\,\mathrm dx\stackrel{u=\frac{3}{x}}{=}-\frac16\int_3^1e^u\,\mathrm du=\frac{e^3-e}{6}. \]

III. Classify the integrand according to its form.

  • If steps 1 or 2 didn't yield a result, based on the form of the integrand \(f(x)\), you should try to figure out which method of integration you should use.
  • Trigonometric functions. If \(f(x)\) is a product of powers of \(\sin x\) and \(\cos x\), of \(\tan x\) and \(\sec x\), or of \(\cot x\) and \(\csc x\), then you should apply trigonometric integration.
    • Example. 7.5.13. Since the integral \(\int\sin^5t\cos^4t\,\mathrm dt\) has an odd power of \(\sin t\), we leave one \(\sin t\) and convert the rest: \[\begin{multline*} \int\sin^5t\cos^4t\,\mathrm dt=\int\sin t(1-\cos^2t)^2\cos^4t\,\mathrm dt=\int\sin t(1-2\cos^2t+\cos^4t)\cos^4t\\ \stackrel{u=\cos t}{=}-\int u^4-2u^6+u^8\,\mathrm du=-\frac{\cos^5t}{5}+\frac{2\cos^7t}{7}-\frac{\cos^9t}{9}+C. \end{multline*}\]

III. Classify the integrand according to its form.

  • Rational functions. If \(f(x)\) is a rational function, we use partial fractions.
    • Example. 7.5.26. \(\int\frac{3x^2+1}{x^3+x^2+x+1}\,\mathrm dx\). Since one can check that -1 is a root of the denominator, we can divide in by \((x+1)\): \(x^3+x^2+x+1=(x+1)(x^2+1)\).
    • Since \(x^2+1\) is an irreducible quadratic polynomial, we need to solve \[ \frac{3x^2+1}{x^3+x^2+x+1}=\frac{A}{x}+\frac{Bx+c}{x^2+1}, \] that is \[\begin{align*} 3x^2+1&=A(x^2+1)+(Bx+C)x\\ &=x^2(A+B)+xC+A. \end{align*}\]
    • We get \(C=0\), \(A=1\), and thus \(B=2\).
    • Therefore, we have \[ \int\frac{3x^2+1}{x^3+x^2+x+1}\,\mathrm dx=\int\frac{1}{x}+\frac{2x}{x^2+1}\,\mathrm dx=\ln|x|+\ln|x^2+1|+C. \]

III. Classify the integrand according to its form.

  • Integration by parts. If \(f(x)\) is a product of a polynomial and a transcendental function (ie.~trigonometric, exponential or logarithmic), then you should use integration by parts.
    • Example. 7.5.15. \(\int x\sec x\tan x\,\mathrm dx\). Since this is a product of a polynomial and a trigonometric function, we can try to derivate the polynomial, and integrate the trigonometric function: \[ \int x\sec x\tan x\,\mathrm dx=x\sec x-\int\sec x\,\mathrm dx=x\sec x-\ln|\tan x+\sec x|+C. \]
    • Example 7.5.3. \(\int\sqrt y\ln y\,\mathrm dy\). We haven't talked about this before: when applying integration by parts to the product to a polynomial and a logarithmic function, unlike the other cases, you can try to integrate the polynomial, and derivate the logarithmic function: \[ \int\sqrt y\ln y\,\mathrm dy=\frac23\left(\left[y^{3/2}\ln y\right]_1^4-\int_1^4\sqrt y\,\mathrm dy\right)=\frac{32}{3}\ln2-\frac49[y^{3/2}]_1^4=\frac{32}{3}-\frac{31}{9}. \]

III. Classify the integrand according to its form.

  • Radicals. If \(\sqrt\pm x^2\pm a^2\) occurs, use the appropriate trigonometric substitution.
    • Example. 7.5.11. \(\int\frac{1}{x^3\sqrt{x^2-1}}\,\mathrm dx\). We see a \(\sqrt{x^2-1}\), so we should substitute \(x=\sec\theta\): \[\begin{multline*} \int\frac{1}{x^3\sqrt{x^2-1}}\,\mathrm dx\stackrel{x=\sec\theta}{=}\int\frac{\tan\theta\sec\theta}{\sec^3\theta\tan\theta}\,\mathrm d\theta=\int\cos^2\theta\,\mathrm d\theta\\ =\frac12\int1+\cos2\theta\,\mathrm d\theta=\frac12\sec^{-1}x+\frac14\sin2\theta+C=\frac12\sec^{-1}x+\frac12\sqrt{1-(\cos^{-1}x^{-1})^2}\cos^{-1}x^{-1}+C. \end{multline*}\]
  • Rationalizing substitution. If a function \(g(x)\) with a radical occurs, you can try \(u=g(x)\).
    • Example. 7.5.23. \(\int_0^1(1+\sqrt x)^8\,\mathrm dx\). Let's try substituting \(u=1+\sqrt x\), which gives \(\mathrm du=\frac12x^{-1/2}\,\mathrm dx\), that is \(2(u-1)\mathrm du=\mathrm dx\): \[ \int_0^1(1+\sqrt x)^8\,\mathrm dx=2\int_1^2(u-1)u^8\,\mathrm du=\left[\frac15u^{10}-\frac29u^9\right]_1^2=\frac{2^{10}-1}{5}-\frac{2^{10}-2}{9}. \]

IV. Try again.

  • If none of the first three steps yielded a solution, remember that there are basically only two methods: substitution and parts.
  • Try substitution. You can basically try various subtitutions.
    • Example. 7.5.21. \(\int\tan^{-1}\sqrt x\,\mathrm dx\). Here, the function with the radical in inside another function. We can still try \(u=\sqrt x\): \[ \int\tan^{-1}\sqrt x\,\mathrm dx=2\int u\tan^{-1}u. \]
    • Since we have \((\tan^{-1}u)'=\frac{1}{1+u^2}\), we can use integration by parts, with integrating \(u\), and derivating \(\tan^{-1}u\): \[ =u^2\tan^{-1}u-\int\frac{u^2}{1+u^2}\,\mathrm du=x\tan^{-1}\sqrt x-\int 1-\frac{1}{1+u^2}\,\mathrm du=x\tan^{-1}\sqrt x-\sqrt x+\tan^{-1}\sqrt x+C. \]
  • Try parts. Recall that there are cases where you can find \(\int f(x)\,\mathrm dx\) via integration by parts with \(u=f(x)\) and \(\mathrm dv=\mathrm dx\).
    • Example. 7.5.14. \[ \ln(1+x^2)\,\mathrm dx=x\ln(1+x^2)-\int\frac{2x^2}{1+x^2}\,\mathrm dx=x\ln(1+x^2)-2x+2\tan^{-1}x+C. \]

IV. Try again.

  • Manipulate the integrand. You can try performing various algebraic manipulations on the integrand, for example expanding it so that you can apply some algebraic or trigonometric formula.
    • Example. \(\int\frac{\cos x}{1-\sin x}\,\mathrm dx\). Since we have \[ \frac{\cos x}{1-\sin x}=\frac{\cos x}{1-\sin x}\frac{1+\sin x}{1+\sin x}=\frac{\cos x+\sin x\cos x}{1-\sin^2x}=\frac{\cos x}{\cos^2x}+\frac{\sin x\cos x}{\cos^2x}=\sec x+\tan x, \] we get \[ \int\frac{\cos x}{1-\sin x}\,\mathrm dx=\int\sec x+\tan x\,\mathrm dx=\ln|\tan x+\sec x|-\ln|\cos x|+C. \]
  • Exercises. I'm sorry, but I'm going to say you should do as many problems as you can from this section. This is one of the most difficult parts of this course. You should get in as much practice as possible.

Can we integrate all continuous functions?

  • One might ask if it is possible to calculate the integral of any continuous function.
    • In Calculus I, you have seen that every continuous function is integrable.
    • So you might ask: is it possible to calculate the integral of every continuous function?
    • Of course, this invites the question: what does it mean to calculate the integral of a continuous function? What does it mean to describe a function?
    • The functions we have been dealing with are called elementary function. An elementary function is one you can obtain via addition, multiplication, and composition from polynomial, exponential, trigonometric functions, and their inverses.
    • So we should rather ask: is the integral of every elementary function an elementary function.
    • The answer is no. The most striking example is that the antiderivative of the *density function of a normal distribution with mean \(\mu\) and standard deviation \(\sigma\): \[ \phi(x)=\frac{1}{\sqrt{2\sigma^2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \] is not an elementary function. This is one of the most commonly used density functions that can create bell curves.