# 4.7: Integration Techniques

**At Grade**Created by: CK-12

## Integration by Substitution

This lesson describes two important tricks for coming up with antiderivatives and integrating. The first could be explained simply as a means of reducing complex looking integrals to simpler more friendly ones. Or, even on a more basic level, substitutions can be used to make complicated looking functions in general look simpler. Consider the function \begin{align*}f(x)=e^{3x+2}\end{align*}

\begin{align*} \frac{du}{dx} = 3 \Rightarrow du = 3 dx \end{align*}

So that:

\begin{align*}e ^ {3x + 2} = 3 e ^ u du\end{align*}

which looks a lot simpler. If we were to integrate the function we would carry out the replacements in the following way:

\begin{align*}\int\limits_{x = a}^{x = b} e ^ {3x + 2} dx = \int\limits_{u = 3a + 2}^{u = 3b + 2} 3 e ^ u du\end{align*}

PROCESS

It is nice to present a difficult problem in detail and then have students work alone or in small groups to solve a similar problem. For example, the following problem could be presented on the board with clear explanations for each step:

Solve the Integral:

\begin{align*} \int\limits_{0}^{\frac{\pi}{4}} \text {Tan}(x) dx\end{align*}

We begin by recognizing that \begin{align*}\mathrm {Tan}(x) = \frac{\mathrm {Sin}(x)}{\mathrm {Cos} (x)}\end{align*} and making the \begin{align*}u-\end{align*} substitution: \begin{align*}u=\mathrm {Cos}(x)\end{align*} so that \begin{align*}du=-\mathrm {Sin}(x)dx\end{align*} and at \begin{align*}x=0,u=\mathrm {Cos}(0)=1\end{align*} and at \begin{align*}x = \frac{\pi}{4}, u = \mathrm {Cos} \left ( \frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}\end{align*}:

\begin{align*}\int\limits_{0}^{\frac{\pi}{4}} \text {Tan} (x) dx = \int\limits_{0}^{\frac{\pi}{4}} \frac{\text {Sin}(x)dx}{\text {Cos}(x)dx} = - \int\limits_{1}^{\frac{\sqrt{2}}{2}} \frac{1}{u} du = \text {Log} \left ( \frac{2}{\sqrt{2}} \right ) = \frac{1}{2} \text {Log} (2) \cong 0.35 \end{align*}

PRODUCTS

Substitutions should be tested by having students perform integrals that are simplified with a substitution. The following are good questions to get them warmed up:

1. Show that \begin{align*}\frac{\mathrm {Sin}(x)}{\mathrm {Cos}(x)} dx = \frac{du}{u}\end{align*} using an appropriate \begin{align*}u-\end{align*}substitution.

2. Show that \begin{align*}\frac{x dx}{\sqrt{1 - x ^ 2}} = u ^ {-\frac{1}{2}} du\end{align*} using an appropriate \begin{align*}u-\end{align*}substitution.

3. Show that \begin{align*}\frac{dx}{\sqrt{4 - x ^ 2}} = du \end{align*} using an appropriate substitution. (hint: This problem will involve the identity that \begin{align*} 1 - \mathrm {Sin} ^ 2 (x) = \mathrm {Cos} ^ 2 (x)) \end{align*}

4. Solve \begin{align*}\int\limits_{1}^{100} \frac{e ^ x}{ e ^ {2x} - 1}dx\end{align*} using an appropriate \begin{align*}u-\end{align*}substitution

## Integration by Parts

CONTENT

Integration by parts should be tried whenever an integral is a product of two functions that each individually have known antiderivatives. In physics this trick is often applied to functions which vanish at the limit points. For example, if we assume that at least one of \begin{align*}f(x)\end{align*} or \begin{align*}g(x)\end{align*} is zero at \begin{align*}x=a\end{align*} and at \begin{align*}x=b\end{align*}, then their product \begin{align*}f(a)g(a)=0\end{align*} and similarly \begin{align*}f(b)g(b)=0\end{align*} so we can see that:

\begin{align*}\int\limits_{a}^{b}f(x) g'(x) dx & = \int\limits_{a}^{b}[ f(x)g(x)] dx - f' (x) g(x) dx \\ & = \cancel{( f(b) g(b) - f(a)g(a))} - \int\limits_{a}^{b} f'(x)g(x) dx \\ & = - \int\limits_{a}^{b}f'(x)g(x)dx \end{align*}

so that integrating by parts allows us to simply “transfer” the derivative from the function \begin{align*}g(x)\end{align*} onto the function \begin{align*}f(x)\end{align*} only by incurring a negative sign.

PROCESS

To teach this it is nice to do a problem in detail and then have students work individually on a similar problem trying to recall your steps as they go. To teach integration by parts put a problem on the board, like:

\begin{align*}\int x e ^ x dx\end{align*}

and present this as a sort of riddle. It may even be fun to have students divide into small groups and think about this problem. Remember, the goal is to come up with some function \begin{align*}f(x)\end{align*} whose derivative is equal to \begin{align*}f'(x) = x e ^ x\end{align*} . Focusing heavily on the additive constant will only frustrate students and take the fun out of the game.

For the problem above, students should be encouraged to literally just guess, try their guess, and then try to fix it. If they were to try \begin{align*}f(x)=x e^x\end{align*} they would find that:

\begin{align*} f'(x) = e ^ x + x e ^ x \end{align*}

So the derivative is almost right, except for that pesky \begin{align*}e^x\end{align*}. Maybe if they subtracted the integral of that. But \begin{align*}\int e ^ x dx = e ^ x\end{align*} so they could just try instead \begin{align*}f(x) = x e ^ x - e ^ x\end{align*}. This function has derivative:

\begin{align*} f'(x) = e ^ x + x e ^ x - e ^ x = x e ^ x \end{align*}

After this the formal equations can be introduced and they are much more likely to be understood than if they are discussed without a concrete example.

PRODUCTS

1. Show that \begin{align*}x ^ 2 e ^ x dx = \frac{d}{dx} ( x ^ 2 e ^ x) dx - \frac{d}{dx} ( 2 x e ^ x) dx + 2e ^ x dx \end{align*} and that therefore:

\begin{align*} \int x ^ 2 e ^ x dx = x ^ 2 e ^ x - 2x e ^ x + 2 e ^ x\end{align*}

2. Show that \begin{align*}\mathrm{log} (x) dx = \frac{d}{dx} ( x \mathrm{log} (x)) dx - 1 dx \end{align*} and that therefore:

\begin{align*}\int \text {log} (x) dx = x \text {log} x - x \end{align*}

3. Find the following integral:

\begin{align*} \int \text{Sin}(x) e ^ x dx \end{align*}

## Integration by Partial Fractions

CONTENT

This lesson describes a general means by which complicated fractions of polynomials, or *rational functions*, can be written out as a sum of simpler fractions. Often one can use a guess and check technique to find the correct partial fraction decomposition. As a very simple example, consider the fraction:

\begin{align*} \frac{x}{x ^ 2 - 1}\end{align*}

We recognize that the denominator is the product \begin{align*}(x-1)(x+1)\end{align*} and so it seems likely that this could arise from adding two fractions with \begin{align*}x-1\end{align*} and \begin{align*}x+1\end{align*}. We might naively look at:

\begin{align*}\frac{1}{ x - 1} + \frac{1}{ x + 1} \end{align*}

However, adding these we obtain:

\begin{align*}\frac{1}{x - 1} + \frac{1}{x + 1} = \frac{2x}{x ^ 2 - 1} \end{align*}

and so clearly multiplying this result by \begin{align*}\frac{1}{2}\end{align*} will give the desired decomposition:

\begin{align*}\frac{\frac{1}{2}}{x - 1} + \frac{\frac{1}{2}}{x + 1} = \frac{x}{x ^ 2 - 1}\end{align*}

PROCESS

The guess and check technique is only of limited value for more complicated fractions and is only possible with some experience. Therefore, it will be important for students to practice with a number of examples. However, it should be clear that the integration part of these problems is really not important to the concept itself of partial fraction decomposition. So it is recommended that students are given the opportunity to gain experience with partial fractions without having to integrate them afterwards.

This can be done with a game. Have students take out only paper and pencil and divide the class into two teams. On the board you can write some rational function like:

\begin{align*}\frac{x + 1}{ x ^ 2 - 2x + 1} \end{align*}

and then call on the student who believes s/he has decomposed this correctly for his/her team. The student will have to present the work on the board, and if it is correct:

\begin{align*}\frac{x + 1}{x ^ 2 - 2x + 1} = \frac{1}{x - 1} + \frac{2}{ ( x - 1) ^ 2} \end{align*}

then the team will earn \begin{align*}1\end{align*} point. However, that person now cannot come to the board for his/her team and another member will have to come the next time the team believes they have a correct answer. They can work as a group on a problem, but the entire team must come to the board at least once before repeating members.

PRODUCTS

Again, it is not important that students complete the integration in these problems to learn and master the technique of partial fractions. Of course, this will eventually be important and should ideally be practiced. However if time is limited then the decomposition itself can be honed.

## Trigonometric Integrals

CONTENT/PROCESS/PRODUCTS

The material presented here is more or less a recipe for various possible combinations of sines, cosines, tangents and secants. Without some motivation, most students will find this lesson to be fairly dull. So it is recommended that a solid example of some kind be used to bring this to life. For example, one might discuss the voltage being in an electrical wall socket. This voltage is actually described by the function:

\begin{align*}V(t) = V_{peak} \text {Sin}( 2 \pi f t) \end{align*}

where \begin{align*}V_{peak}\end{align*} is called the peak voltage and \begin{align*}f\end{align*} is the frequency. You may know that sockets in the United States give \begin{align*}120 \text {Volts}\end{align*} but this is actually the so-called “root-mean squared” or *rms* value.

To understand this it is important to have a look at the plot for voltage above:

The voltage fluctuates in time between the same magnitudes and therefore averages to zero over any given cycle. Given that the period of a cycle is \begin{align*}T = \frac{1}{f}\end{align*} we have:

\begin{align*} V_{ave} = \frac{1}{T} \int\limits_{0}^{T} V(t) dt = \frac{V_{peak}}{T} \int\limits_{0}^{\frac{1}{f}}\text {Sin} ( 2 \pi f t) dt = \frac{V_{peak}}{T} \frac{1}{2 \pi f} [ - \text {Cos} ( 2 \pi f t)] _{0} ^ \frac{1}{f} = 0 \end{align*}

So the average voltage produced by any wall-socket is actually zero!

What we really want instead is a measure of how big the voltage is on average, and one way to do that is to make it positive everywhere and then take the average. This could be done with absolute values, but the simpler way is by squaring the function, taking its average, and then taking the square root of the result:

\begin{align*}V_{rms} = \frac{1}{T} \sqrt{\int\limits_{0}^{T} [V(t)] ^ 2 dt} = \frac{V_{peak}}{T} \sqrt{ \int\limits_{0}^{T} \text {Sin} ^ 2 ( 2 \pi f t ) dt}\end{align*}

This function will not have an average of zero since it is everywhere positive:

and in fact we need to use integration of a power of sine here to obtain:

\begin{align*}V_ {rms} = \frac{V_{peak}}{T} = \sqrt{\int\limits_{0}^{T} \text {Sin} ^ 2 ( 2 \pi f t ) dt} = \frac{V_{peak}}{T} \sqrt{ \left [ \frac{t}{2} - \frac{\text {Sin}[4 f \pi t]}{8 f \pi} \right ] _{0} ^ {T}} = \frac{V_{peak}}{\sqrt{2}} \end{align*}

## Trigonometric Substitutions

CONTENT

Trig substitutions arise as useful tools when integrals defined in Cartesian Coordinates are actually more effectively evaluated in a different coordinate system. However, this cannot be discussed in detail without going into \begin{align*}2 \mathrm {dimensions}\end{align*}. So it should suffice to say that students should look to perform a trig substitution when the variable \begin{align*}x\end{align*} is involved in a term that looks like one of the trig identities.

For example, if there is a part of the integral that involves a term like:

\begin{align*} a ^ 2 - x ^ 2 \end{align*}

then one might notice that if we set \begin{align*}x=a \mathrm {Cos}(u)\end{align*} this will become:

\begin{align*}a ^ 2 - x ^ 2 = a ^ 2 - a ^ 2 \text {Cos} ^ 2 (u) = a ^ 2 \text {Sin} ^ 2 (u)\end{align*}

or equivalently if we set \begin{align*}x=a \mathrm {Sin}(u)\end{align*} this will become:

\begin{align*}a^2-x^2=a^2-\text {Sin}^2 (u)=a^2 \text {Cos}^2 (u)\end{align*}

Similarly, if a part of the integral involves a term like:

\begin{align*}a^2+x^2\end{align*}

then one might notice that if we set \begin{align*}x=a \text {Tan}(u)\end{align*} this will become:

\begin{align*}a^2+x^2=a^2+a^2 \text {Tan} ^2 (u)=a^2 \text {Sec}^2 (u)\end{align*}

PROCESS/PRODUCTS

It may be best to review quickly where the trig identities that are used in these substitutions come from. If students are encouraged to maintain a picture of a triangle in their minds, then it will not be difficult to recall each identity:

Then the teaching may be done by carrying out a few detailed calculations and requiring students to perform similar work just after. They should try to focus on your logic in each step instead of taking notes or memorizing formulas. That way when they attack the problems on their own the struggle to recall your logic will make the knowledge longer lasting.

## Improper Integrals

CONTENT

Improper integrals can be described as simply a two-step process: We perform the integrals for values that we can do, and then look at the limit as we approach the values we’re not sure about. When a limits itself is infinite, we replace that limit by an arbitrary letter and then after we are done we let that letter go to infinity. When the integral passes over a point of infinite discontinuity then we simply replace that point by an arbitrary letter and then look at its limit after solving the integral again.

PROCESS

To teach this it is nice to give a geometric presentation. We would like to know if the area beneath the curve \begin{align*}f(x) = \frac{1}{x}\end{align*} is finite:

The integral performed in the text shows us that the area does not converge. However, if we look at the area beneath \begin{align*}f(x) = \frac{1}{x ^ 2}\end{align*} instead:

we find that it does converge to a finite number. Points of infinite discontinuity can be treated in a similar way.

PRODUCTS

A variety of different and interesting integrals can be performed to demonstrate improper limits. Most notably students may be asked to recall probability densities \begin{align*}p(x)\end{align*}. Since the total probability of any \begin{align*}x\end{align*} occurring must be identically \begin{align*}1\end{align*}, we should have:

\begin{align*} \int\limits_{- \infty}^{\infty} f(x) dx = 1\end{align*}

Students could be asked to solve an equation like this where \begin{align*}f(x)\end{align*} is some unnormalized probability density like, say, \begin{align*}f(x) = C e ^ {-|x|}\end{align*}:

By solving the equation students would be finding the constant \begin{align*}C\end{align*} so that \begin{align*}f(x)\end{align*} is a valid probability density:

\begin{align*}1 = C \int\limits_{- \infty}^{\infty} e ^ {-|x|} dx = 2C \int\limits_{0}^{\infty} e ^ {-x} dx = 2C[ 1 - 0] = 2C\end{align*}

So clearly \begin{align*}C = \frac{1}{2}\end{align*} will work and we should have\begin{align*} f(x) = \frac{1}{2} e ^ {-|x|}\end{align*}.

As a general rule, giving students examples that have concrete applications will make this material more interesting. Students who may not be as motivated as others will appreciate a little less abstraction wherever possible.

## Ordinary Differential Equations

CONTENT

Notice that the right hand side of the equation for a Linear ODE:

\begin{align*}Y ^ {(n)} = \Sigma a _ i (x) Y ^ {(i)} + r(x) \end{align*}

is actually a function of two variables, \begin{align*}y\end{align*} and \begin{align*}x\end{align*}. Therefore, the general analysis of any ODE actually belongs in a multivariable calculus class.

The technique of sketching slope fields to visualize the isoclines of for simple differential equation \begin{align*}\frac{dy}{dx} = F (x ,y)\end{align*} is a very powerful one. Students will be interested to know that a great many problems in math and science reduce to some equation of the form \begin{align*}\frac{dy}{dx} = F (x ,y)\end{align*} that cannot be solved exactly. We therefore aim instead to create better and better approximations of the solutions using the slope fields.

PROCESS

In teaching this topic students are likely to feel a little confused about the complicated looking equations and all the indices in the approximation techniques. To put them at ease, it is nice to have at least one very clearly presented and simple example. The first example should help build intuition by describing a particular equation, like \begin{align*}\frac{dy}{dx} = xy\end{align*}, in excruciating detail with a careful plot of the slope fields and isoclines:

A discussion on the numerical methods for solving the differential equations follows naturally from a good example. Before actually applying some analytic trick it is a good idea to see where certain points take you along the numerical approximations. This can be very effectively done by tracking your progress along a plot of the isoclines like above. One numerical solution will generally not follow a single isocline, but as long as the step-size is small and the starting point is not near any major singularities, it should very closely follow the contour of an isocline.

PRODUCTS

Students can be tested on this material by being given simple differential equations to solve that closely mimic problems already solved in the text or in class. They should be encouraged to work in groups and to look for problems that look similar. In following the work and changing it as needed they will become experts at simple ODEs in no time at all.

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