Other Strategies for Integration

Section 3.5 Other Strategies for Integration

References.

OpenStax Calculus Volume 2, Section 3.5 ¹.
Calculus, Early Transcendentals by Stewart, Section 7.6.

Strategy for Evaluating Indefinite Integrals.

Here is a basic strategy for combining the various integration methods from this Chapter and Chapter 1. After any step that does not give the final answer, start again from the top on each piece.

Use known integrals (basic anti-drivatives): build a list of them in your notes. Also use lists of integrals like those in Appendix A of OpenStax Calculus Volume 2 ², which gives 113 such integrals.
Simplify with algebra, trig., and such, break up into a sum of simpler integrals, and move any constant common factors outside an integral.
Try Substitution, \(u=g(x)\text{:}\) often \(u\) is the term inside a composition. Do not forget to convert the differential too: \(du=g'(x) dx\text{.}\)
Try Integration by Parts, \(\int u \, dv = uv - \int v \, du\text{.}\) When arranging as \(\int u \, dv\text{,}\) try to put as much as possible into \(dv\) while still being able to integrate it to get \(v=\int dv\text{.}\)
Finally, some important special cases:
1. For functions involving the square root of a quadratic, or powers of such a root and powers of \(x\text{,}\)try the inverse trigonometric substitutions of Section 3.3.
2. For products of trigonometric functions, try the special substitutions and half-angle formulas of Section 3.2.
3. For rational functions, use partial fractions expansions (and synthetic division if necessary), as in Section 3.4.

Using Reduction Formulas.

When using tables of integrals such as in Appendix A of OpenStax Calculus Volume 2 ³ one very useful and flexible case is Reduction Formulas, which express an integral in terms of a somehow simpler integral, and which are typically used repeatedly or in combination until the original integral is in terms of one that can be evaluated directly.

This reduction process is seen with integration by parts of functions like \(x^3 e^x\text{.}\)

Using \(dv=e^x dx\text{,}\) \(u=x^3\text{,}\) one gets \(\ds\int x^3 e^x \, dx = x^3 e^x - 3 \int x^2 e^x \, dx\) which gives a very similar integral where integration by parts with \(dv=e^x dx\) again gives \(\ds \int x^2 e^x \, dx = x^2 e^x - 2 \int x e^x \, dx.\)

Yet another integration by parts with \(dv=e^x dx\) gives \(\ds \int x e^x \, dx = x e^x - \int e^x \, dx.\)

Finally the integral needed is a known one:

\begin{equation} \int e^x \, dx = e^x + C.\tag{3.5.1} \end{equation}

Putting all these together,\(\ds \int x^3 e^x \, dx = x^3 e^x - 3 x^2 e^x + 6 x e^x - 6 e^x + C.\)

But the three integrations by parts above are very similar, so doing them all seems redundant and would be far more so if we needed to evaluate something like \(\int x^{17} e^x \, dx\text{.}\) Instead we can roll all these integrations into one more general Reduction Formula:

\begin{equation} \int x^n e^x \, dx = x^n e^x - n \int x^{n-1} e^x \, dx\tag{3.5.2} \end{equation}

which can be verified by a single integration by parts with \(dv=e^x \, dx\text{,}\) \(u=x^n\text{.}\)

Using this successively with \(n=3\text{,}\) then \(n=2\text{,}\) then \(n=1\) and then using (3.5.1) gets the above result.

In fact, this formula is one of the reduction formulas in that Appendix A ⁴: number 43, though there it is in a slightly more general form

\begin{equation*} \int u^n e^{au} \, du = \frac{1}{a} u^n e^{au} - \frac{n}{a} \int u^{n-1} e^{au} \, du \end{equation*}

so we also need to specify \(a=1\) when using it. (We could even do the final step using number 3 from there.)

Any indefinite integral that can be evaluated in terms of elementary functions can be reduced to using a combination of formulas in that list through Substitution, Integration by Parts, and various algebraic and trigonometric simplifications like those seen in this Chapter.

Evaluating Definite Integrals.

Often the best approach for a definite integral \(\int_a^b f(x) \, dx\) is to first evaluate the indefinite integral \(F(x) + C\) and then use the Fundamental Theorem of Calculus:

\begin{equation*} \int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a). \end{equation*}

However with substitutions and inverse substitutions, it sometimes saves effort to convert the definite integral to the new variable, to avoid having to convert back to the original variable:

Evaluating Definite Integrals with Substitutions.

For substitution \(u=g(x)\)

\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*}
For inverse substitution \(x=h(t)\)

\begin{equation*} \int_{x=a}^{x=b} f(x) \, dx = \int_{t=c}^{t=d} f(h(t)) h'(t) \, dt, \end{equation*}

where now you have to solve equations to get \(c\) from \(a\) and \(d\) from \(b\text{:}\) \(h(c)=a\text{,}\) \(h(d)=b\text{.}\)

Section Study Guide.

Read the first part of Calculus Volume 2, Section 3.5 ⁵, about Tables of Integrals.

There is only one worked example there, so work a selection from Exercises in each of the ranges 240–260, 279–284, and 285–288.

openstax.org/books/calculus-volume-2/pages/3-5-other-strategies-for-integration

openstax.org/books/calculus-volume-2/pages/a-table-of-integrals