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Integrals involving trig substitutions

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When is it appropriate to solve an integral with a trig substitution? First of all, keep in mind that a trig substitution doesn’t always work. Even when it does work, you are often left with an integral that will require other techniques such as a u substitution or integration by parts. But if you are willing to put in a little effort (and you know your trig identities), trig substitutions allow you to find the antiderivatives of some rather complicated functions.

There are three conditions that you look for—each a radical term of a particular form in the integrand. Each condition is associated with a different substitution. After you make the substitution, you simplify the integrand and go from there.

Term

Substitution

Radical becomes…

\sqrt{a^2-x^2} \text{Let } x=a \sin \theta; \, dx=a \cos \theta \,d \theta a \cos \theta
\sqrt{a^2+x^2} \text{Let } x=a \tan \theta; \, dx=a \sec^2 \theta \,d \theta a \sec \theta
\sqrt{x^2-a^2} \text{Let } x=a \sec \theta; \, dx=a \sec \theta \tan \theta \,d \theta a \tan \theta

 

Before we look at some example integrals, let’s see why the first radical term above simplifies to a \cos \theta. It’s pretty straightforward if you know your trig identities:

\sqrt{a^2-x^2}=\sqrt{a^2-(a \sin \theta)^2}= \sqrt{a^2-a^2 \sin^2 \theta}= \sqrt{a^2(1- \sin^2 \theta)} =

 …….. \sqrt{a^2 \cos^2 \theta} = a \cos \theta

Example. Integrate the following:

\text{a.} \displaystyle \int \frac{dx}{\sqrt{1-x^2}} \qquad \text{b.} \int \frac{x^3}{8 \sqrt{4+x^2}}dx \qquad \text{c.} \int x \sqrt{x^2-4} \; dx

Solutions:

  1. (Does this integral look familiar?) Here, a=1, so use x= \sin \theta. Using the first line of the table above:
    …..
    \displaystyle \int \frac{dx}{\sqrt{1-x^2}}= \int \frac{\cos \theta \, d \theta}{\cos \theta}= \int d \theta = \theta +C
    …..
    But since x= \sin \theta, \theta=\sin^{-1} x.
    …..
    \therefore \displaystyle \int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1} x+C
  2. Here, a=2 and we use line 2 from the table above (x=2 \tan \theta). Note that x^3=8 \tan^3 \theta. Upon substitution,
    …..
    \displaystyle \int \frac{x^3}{8 \sqrt{4+x^2}}\, dx= \int \frac{8 \tan^3 \theta}{8(2 \sec \theta)}2 \sec^2 \theta \, d \theta = \int \tan^3 \theta \sec \theta \, d \theta
    …..
    Hmm. This is going to take a little bit of extra work… Time to pull out some trig identities:
    …..
    \int \tan^3 \theta \sec \theta \, d \theta = \int \tan^2 \theta \tan \theta \sec \theta \, d \theta = \int (\sec^2 \theta -1) \tan \theta \sec \theta \, d \theta
    …..
    Now a u substitution, letting u= \sec \theta:
    …..
    \displaystyle \int (u^2-1) \, du= \frac{1}{3} u^3-u+C= \frac{1}{3} \sec^3 \theta - \sec \theta +C
    …..
    How do we get our answer back in terms of x? Draw a triangle that shows how x and \tan \theta are related, then use the Pythagorean theorem to find an expression for \sec \theta. In the triangle below,
    …..
    x=2 \tan \theta \therefore \sec \theta = \dfrac{\sqrt{4+x^2}}{2}
    Substitute into the integral above to get:
    \displaystyle \int \frac{x^3}{8 \sqrt{4+x^2}} dx=\frac{1}{3} \sec^3 \theta - \sec \theta +C= \frac{(4+x^2)^{3/2}}{24}- \frac{(4+x^2)^{1/2}}{2}+C

    This can be simplified further by factoring:
    …..
    \displaystyle \int \frac{x^3}{8 \sqrt{4+x^2}}dx= \frac{(4+x^2)^{3/2}}{24}- \frac{(4+x^2)^{1/2}}{2}+C= \frac{1}{24}(x^2-8) \sqrt{4+x^2}
    …..
    Whew!

  3. This is a trick question. Even though it fits the condition given in the table (and you could integrate with a trig substitution if you wanted), it’s easier to do this one with a u substitution: u=x^2-4 and du=2x \; dx:
    …..
    \displaystyle \int x \sqrt{x^2-4} \, dx= \frac{1}{2} \int \sqrt{u} \, du= \frac{1}{3}u^{3/2}+C= \frac{1}{3}(x^2-4)^{3/2}+C

The lesson here is to look for u substitutions before you look for trig substitutions.

Using f”(x) to interpret f’(x)

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You know that the second derivative of a function is used to characterize the concavity of the function. But did you know that the second derivative also gives you information about the first derivative? Well, of course it does, because the second derivative is the first derivative of the first derivative. To put it simply, when the second derivative is positive, that means the first derivative is increasing. When the second derivative is negative, the first derivative is decreasing.

But wait, I thought when the second derivative is positive, that means the original function is concave up! Well, yes, that’s true too. So that means wherever a function is concave up, its first derivative is increasing. And wherever a function is concave down, the first derivative is decreasing.

This is another way you can analyze the behavior of a function.

u substitutions with definite integrals

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When a definite integral requires a u substitution to solve, be sure to substitute for the limits of integration as well. This way, you don’t need to substitute back in for the original function. Instead, you evaluate the integral using the new (u) limits. Here’s an example to show how this works.

Evaluate:

\displaystyle \int_0^{\pi/4} \sec^2 \theta \tan^2 \theta \; d \theta

This is an obvious candidate for a u substitution. (See other posts on this website for more information on when to use u substitutions.)

Let u= \tan \theta. Then du=\sec^2 \theta \; d \theta.

But don’t stop there! Use your expression for u to determine the new limits as well.

\theta =0 \rightarrow u=0; \; \theta= \dfrac{\pi}{4} \rightarrow u=1

So the new integral becomes

\displaystyle \int_0^{\pi/4} \sec^2 \theta \tan^2 \theta \; d \theta =\int_0^1 u^2 \; du= \left. \dfrac{1}{3}u^3 \right |_0^1=\dfrac{1}{3}-0=\dfrac{1}{3}

You have found the solution to the original integral without needing to put the integral back in terms of \theta.

How to approach integration by parts

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In an earlier post, I described a strategy for approaching integrals and how to decide what technique to use. One question I suggested you ask yourself is “Can I use integration by parts?” In this post, I describe how you decide when integration by parts is the right approach, and how you decide which term will be u and which will be dv.

There is no product rule for integrals, but when you see an integrand that is the product of two functions [and u substitution has been ruled out], integration by parts is often the right approach. For example, each of the following integrals should be solved using integration by parts:

 \displaystyle \int x^2e^x \; dx \\ \displaystyle \int e^x \sin x \; dx \\ \displaystyle \int x \tan^{-1} x \; dx

Once you decide that integration by parts is the correct technique, does it matter which factor you make u and which you make dv? Usually, yes. But don’t worry if you make the wrong choice—you’ll know that pretty quickly. Remember that the purpose of integration by parts is to take a complicated integrand and make it simpler. If you choose your u and dv incorrectly, your integrand will get more complicated. When you find your integral getting worse, just start again and switch your choices for u and dv. (If switching choices doesn’t make your integral look better, perhaps integration by parts isn’t the proper technique for the integral.)

Here’s a trick to help you decide which factor should be the u term: LIPET. This is a mnemonic to help you determine the priority for assigning factors to be u. The acronym stands for the following:

L: logarithmic functions
I: inverse trigonometric functions
P: polynomials
E: exponential functions
T: trigonometric functions

For example, the first integral above is the product of a polynomial and an exponential function. P precedes E in LIPET, so make x2 the u term and ex the dv term. The third integral above is the product of a polynomial and an inverse trig function, so let tan-1 x be the u term and x will be the dv term.

Finally, note that even though integration by parts creates a simpler integral, it won’t always create an integral you can evaluate by inspection. Often, you will need to use another technique to integrate the new integrand, or perhaps you will need to do integration by parts a second time.

Strategies for finding integrals

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How does one become an “expert integrator”? There are a lot of integration techniques, and recognizing which technique is the best is the first step to finding the integral. Many new calculus students look at an integral and don’t even know where to start. Here is a strategy you can adopt when faced with an integral. Carry on this internal dialog with yourself:

  1. Is this something that I can integrate by inspection?
  2. Is this an integral that can be turned into an elementary integral by simplifying?
  3. Is there a trig identity that will simplify the integral?
  4. Is it a candidate for a u substitution?
  5. Can I use integration by parts?
  6. Will decomposition by partial fractions be useful?
  7. Does the integral involve a product of trig functions?
  8. Would a trig substitution be effective?
  9. Can I find it in a table of integrals?

Yes, this is a long list. But with practice, you can work your way through it pretty quickly. [If you are taking Calc AB this year, you will only learn the first four or five steps.] Think of each integration technique as a tool for your toolbox; with each integral you encounter, you decide which tool is the best for that problem.

Let’s look at a couple of examples to see how this works.

Example 1. Evaluate:

\displaystyle \int {\sec}^2 x+x^2-2x \; dx

This is an integral consisting of only elementary derivatives, so you can integrate it by inspection:

\displaystyle \int {\sec}^2 x+x^2-2x \; dx = \tan x+ \dfrac{1}{3}x^3-x^2+C

Note that you need to know the derivatives of the elementary functions in order to recognize when you can integrate by inspection. This means the derivatives of the trig functions, exponential and log functions and the inverse trig functions, not just polynomials.

Example 2. Evaluate:

\displaystyle \int (x^2-1)(2x^2+3x) \; dx

This is not an elementary derivative, so it cannot be integrated by inspection. But the integrand can by simplified by multiplying the terms out, turning it into an elementary derivative:

\displaystyle \int (x^2-1)(2x^2+3x) \; dx \displaystyle \int 2x^4+3x^3-2x^2-3x \; dx=\dfrac{2}{5}x^5+\dfrac{3}{4}x^4-\dfrac{2}{3}x^3- \dfrac{3}{2}x^2+C

Example 3. Evaluate:

\displaystyle \int \tan^2 x \; dx

This is not an elementary derivative. Is there a trig identity that will help? Yes, use the following Pythagorean identity to turn this into a simple integral:

\displaystyle \int \tan^2 x \; dx = \displaystyle \int \sec^2 x-1 \; dx = \tan x-x+C

At this point, you might be thinking “how am I supposed to see that identity?” Well, there’s no substitute for experience and practice. It’s important to know your basic trig identities, so you can rewrite a trig integral in a different form. If one identity doesn’t work, try another. Finding integrals is as much an art as it is a science. Be creative and be inventive.

(Look for other tips on integral problems on this website for strategies on u substitution, integration by parts, products of trig functions and trig substitutions.)

Sum( and Seq( commands on your calculator

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Can you find the sum of the following series?

\displaystyle \sum_{i=1}^{15}\dfrac{2n+1}{3n-2}

 This is neither an arithmetic nor a geometric series, so you don’t have a formula for it. This would be a tedious problem to do by hand. Fortunately, your graphing calculator can do these problems quickly and efficiently.

There are two functions you need to use on your calculator. The seq( command creates a sequence of terms based on a rule that you give. The sum( command adds together the terms in a sequence. Both functions are found on the LIST menu on your calculator. The seq( command is on the OPS submenu and the sum( command is on the MATH submenu.

To sum a series, you combine the two commands. If you have the new operating system on your calculator, it will prompt you for the entries when you select the seq( command. If you have the old operating system, you need to know the syntax for the command. The syntax for the series above is:

 sum(seq((2x + 1)/(3x – 2),x,1,15))

Note that the seq( command has four parameters in the parentheses. From left to right, these are 1) the rule for the nth term of the sequence; 2) the variable name; 3) the first value of the variable; and 4) the final value of the variable. Now all you need to do is type this in to your calculator and let it do the crunching:

 

The “parent” functions

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Do you know your parent functions? I’m surprised every year when some of my students don’t know how to graph some of the elementary functions they’ve learned in class. Your chances for success in precalc and calculus are significantly better if you memorize the shapes of these ten functions. No excuses, no shortcuts. Just learn them.

 y = xn (Power functions)

 

 

 

 

 

 

 

 

 

 

y = |x| (absolute value function)

 

 

 

 

 

 

 

 

 

 

y = ⌊x⌋ (Greatest integer function)

 

 

 

 

 

 

 

 

 

 

y = √x

 

 

 

 

 

 

 

 

 

 

y = 1/x

 

 

 

 

 

 

 

 

 

y= bx (exponential function – growth and decay)

 

 

 

 

 

 

 

 

 

y = log x

 

 

 

 

 

 

 

 

y = sin x and y = cos x

 

 

 

 

 

 

 

 

y = tan x

 

Simplifying “3-stack” and “4-stack” fractions

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I had a physics student a number of years ago who worked a complicated problem and ended up with the following (I’ve changed the actual numbers to make our work here easier to follow):

\dfrac{(2)(3)}{4}=\dfrac{(5)(6)(7)B}{(8)(9)}

Of course, his next step was to solve for B. It should be clear to you, as it was to him, that he needed to multiply both sides by 8 and 9 and divide by 5, 6 and 7. That is what he did. But this is how he wrote the solution:

B= \dfrac{9 \cdot \left( \dfrac{ \left(\dfrac{ 8 \cdot \left( \dfrac{(2)(3)}{4} \right)}{5} \right)}{6} \right)}{7}

His expression was totally correct, and he found the correct value of B, but he made the problem so much harder than he needed to. He created a fraction with five different stacks in it and he needed all those parentheses to keep track of which number was a numerator and which was a denominator.

This is an extreme case of what I see so many students do: they create “3-stack” and “4-stack” fractions all the time when they are simplifying problems. Let’s look at two examples. I’ll solve them with “bad” solutions and then with much smarter solutions (I don’t simplify the answers here because I want to focus on the first step of the solution):

Example 1. Solve:

3x= \dfrac{2}{5}

Bad solution:

x= \dfrac{\dfrac{2}{5}}{3}

 

Smarter solution:

x= \dfrac{2}{5 \cdot 3}

Example 2. Solve:

\dfrac{4}{7}x=\dfrac{2}{5}

 Bad solution:

x= \dfrac{\dfrac{2}{5}}{\dfrac{4}{7}}

 Smarter solution:

x= \dfrac{2 \cdot 7}{5 \cdot 4}

You can see that the smarter solutions are simpler to read and easier to simplify.

It is easy to train yourself to write answers as “2-stack” fractions if you remember one simple rule:

Respect the vinculum.

Um, respect the what?!?

The vinculum. When you write a fraction, the horizontal line that separates the numerator from the denominator is called the vinculum. No one ever talks about it, but it’s a very powerful symbol. It tells you to multiply by every number that is above it and divide by every number that is below it. And it’s as easy as that. So when you are simplifying expressions like the ones in the two examples above or the more complicated example at the beginning of this post, all you need to do is to put numbers that are multiplied above the vinculum and numbers that are divided below the vinculum. And what happens if you are multiplying or diving by a fraction? Then put the numerator on top and the denominator on the bottom if you are multiplying. Flip the fraction over first if you’re dividing (as I did in Example 2). When you do this, you will end up with a 2-stack fraction that can be easily evaluated. Let’s look at the original problem again:

\dfrac{(2)(3)}{4}=\dfrac{(5)(6)(7)B}{(8)(9)}

To solve for B, you will multiply by 8 and 9, so they go on top. And you will divide by 5, 6 and 7, so they go on the bottom. In one step, you’ve solved for B as follows:

B= \dfrac{2 \cdot 3 \cdot 8 \cdot 9}{5 \cdot 6 \cdot 7}

Compare that to the monstrosity at the beginning. So much easier!

P.S. You don’t really have to remember the name “vinculum”. Most people don’t know what the line is called and don’t care. Chances are good your math teacher doesn’t even know the term. To make things even more bizarre, when you write a fraction with a slash instead of a horizontal line like this—2/3 – the slash is called a “virgule”.  Most people don’t know that one and don’t care about it either. All you have to remember is that the horizontal line in a fraction tells you to multiply on top and divide on the bottom.

The Most Common Factoring Mistake

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I am surprised (and a little disappointed) every year when one of my students tries to simplify a polynomial fraction by cancelling out terms that can’t be cancelled out. For example, when faced with

\dfrac{3x+7}{2x-5}

inevitably, a student will ask me, “Can I cancel out the x’s like this?”

 

 

Tears well up in my eyes as I explain that no, the x’s do not cancel. I explain patiently why the x’s do not cancel. And very often, the next time I work with that student, he or she will try to cancel out the x’s again. This is the most common factoring mistake I see students make, and it’s not limited to Algebra students. I’ve even seen Calculus student make this error. That usually makes me sob quite loudly.

If you would like to keep me from crying, then you need to learn how to simplify polynomial fractions. It’s quite simple once you understand that terms that are added do not cancel out. Only factors that are multiplied together can cancel. Let’s start by looking at a fraction with numbers and no variables.

\dfrac{210}{462}

 Can this be simplified? Of course. Most students will divide the top and bottom by 2, then by 3, and then by 7, as follows:

\dfrac{210}{462}=\dfrac{105}{231}=\dfrac{35}{77}=\dfrac{5}{11}

 This is correct. But why can you cancel out a 2 and a 3 and a 7? It’s because they are factors of the numerator and the denominator. Let’s do the same problem by completely factoring the top and bottom first:

\dfrac{210}{462}=\dfrac{2\cdot 3 \cdot 5 \cdot 7}{2 \cdot 3 \cdot 7 \cdot 11}=\dfrac{5}{11}

 

When you write it out like this, you can see that the 2’s cancel, the 3’s cancel, and the 7’s cancel. And they cancel only because they are factors.

Now let’s try to simplify some polynomial fractions. Start by factoring the numerator and denominator completely, then any like factors will cancel:

\dfrac{2x-4}{6x+14}=\dfrac{2(x-2)}{2(x+7)}=\dfrac{x-2}{x+7}

\dfrac{4x+8}{12x+24}=\dfrac{4(x+2)}{12(x+2)}=\dfrac{4}{12}=\dfrac{1}{3}

\dfrac{x^2-1}{x^2+4x+3}=\dfrac{(x+1)(x-1)}{(x+1)(x+3)}=\dfrac{x-1}{x+3}

 

Related Rates: Substituting for extra variables

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The most complicated type of related rate problems for many students is when there is a variable and you have no information about its rate of change. In these problems, you will know how the variable is related to another variable with a known rate. You can then substitute the variable with the known rate in for the variable with the unknown rate. It sounds difficult, but with a little practice, you can master it.

Example: A water tower in the shape of an inverted cone is being filled at a rate of 2 cubic feet per minute. The tower is 10 feet tall and the radius of the base is 5 feet. How fast is the depth of the water changing when the depth is 4 feet?

Solution: First, we draw a picture. This drawing shows a cross section of the water tower, filled to a height of h.

                       

 

 

 

 

 

 

 

 

The variables we know and seek are the following:

\dfrac{dV}{dt}=2; \; \dfrac{dh}{dt}= \, ?

The variables suggest that the equation we need is the volume of a cone:

V= \dfrac{1}{3} \pi r^2h

If we take the derivative now, we will end up with a dr/dt term but we are not given any information about how fast the radius is changing. We get around this problem by solving for r in terms of h and substituting. How are r and h related? Notice that the triangle of water in the drawing above is similar to the triangle of the whole water tower. This means we can set up the following relationship:

\dfrac{5}{10}= \dfrac{r}{h} \rightarrow r= \dfrac{h}{2}

No matter how deep the water is at any time, the radius of the water surface will always be half the height of the water column. Substituting into our equation and simplifying:

V= \dfrac{1}{3} \pi r^2h \rightarrow V= \dfrac{1}{3} \pi \left ( \dfrac{h}{2} \right )^2h \rightarrow V= \dfrac{1}{12} \pi h^3

Now that r has been eliminated, we can differentiate:

\dfrac{dV}{dt}=\dfrac{\pi}{4}h^2 \dfrac{dh}{dt}

Plugging in values gives:

2= \dfrac{\pi}{4}(4)^2 \dfrac{dh}{dt}, \qquad \text{so } \dfrac{dh}{dt}= \dfrac{1}{2 \pi} \, ft/min

(Look for other tips on solving related rate problems on this website if you need help with other aspects of related rate problems.)

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