Mixed Integration Practice

This web exercise gives practice with a wide variety of antidifferentiation problems.

For the convenience of the reader, a review of antiderivatives and the indefinite integral is given.

Review of Antiderivatives

Roughly, antiderivatives ‘undo’ derivatives. Here's a contrived example to illustrate:

• Start with (say) $\,F(x) = x^3\,.$
• Differentiate, giving $\,F'(x) = 3x^2\,.$

• Now, ask the question:  Do we know a function that has derivative $\,3x^2\,$?

  With this question, we're trying to ‘go backwards’ or ‘undo’ the differentiation that we just did.

• Our answer is of course yes! The function $\,x^3\,$ has derivative $\,3x^2\,.$

  We call $\,x^3\,$ an antiderivative of $\,3x^2\,,$ since $\,x^3\,$ is a function that has derivative $\,3x^2\,.$

• Of course, the function (say) $\,x^3 + 7\,$ also has derivative $\,3x^2\,,$ since the derivative of a constant is zero: $$ \begin{align} \frac{d}{dx}(x^3 + 7) &= \frac{d}{dx}(x^3) + \frac{d}{dx} (7)\cr\cr &= 3x^2 + 0\cr\cr &= 3x^2 \end{align} $$

  Thus, $\,x^3 + 7\,$ is also an antiderivative of $\,3x^2\,.$

• Indeed, the entire family of functions $\,x^3 + C\,,$ one function for each real number $\,C\,,$ are antiderivatives of $\,3x^2\,.$

  Notice that the functions in the family $\,x^3 + C\,$ all have the same shape; they differ only by moving up/down.

More precisely, we have:

Thus, an antiderivative of $\,f\,$ is a function that has derivative $\,f\,.$

If you're able to find a single antiderivative of $\,f\,,$ call it $\,F\,,$ then there are an infinite number of antiderivatives, each of the form: $$F(x) + C$$ Here, $\,C\,$ represents any real number, and is called an arbitrary constant.

The process of finding antiderivative(s) is called antidifferentiation.

Review of the Indefinite Integral

The symbol $$\int f(x)\,dx$$ is called the indefinite integral of $\,f\,,$ and represents all the antiderivatives of $\,f\,.$ Thus, the symbol $\,\int f(x)\,dx\,$ represents a family of functions.

Reading an Indefinite Integral Aloud

Remember that mathematics is primarily a written language, not a spoken language. Most of the time that you work with $\,\int f(x)\,dx\,,$ you'll be looking at it.

However, you may sometimes need to read ‘$\,\int f(x)\,dx\,$’ aloud. I've heard all of these variations in reading $\,\int f(x)\,dx\,$ aloud (and more):

• the integral of $\,f\,$ of $\,x\,$ with respect to $\,x$
• the integral of $\,f\,$ of $\,x$
• the integral of $\,f\,$ of $\,x\,$ dee $\,x$

You can say ‘indefinite integral’ instead of just ‘integral’ in any of these verbalizations, if desired for emphasis.

Using Different Dummy Variables in an Indefinite Integral

Using different dummy variables, the indefinite integral of $\,f\,$ can be alternatively written as

Continuing the prior example, we can write:

The arbitrary constant is typically denoted by $\,C\,$ or $\,K\,.$ Be sure to include the arbitrary constant, or else you're only giving one antiderivative, instead of the entire infinite family of antiderivatives!

There's a Definite Integral, Too

Later on (in Chapter 7 of this Calculus course) we'll study the definite integral of $\,f\,$ on $\,[a,b]\,,$ which is denoted by:

Note the limits of integration $\,a\,$ and $\,b\,.$ These limits of integration are what distinguishes a definite integral from an indefinite integral.

Notation for Integrals

Both $\,\int f(x)\,dx\,$ (an indefinite integral) and $\,\int_a^b f(x)\,dx\,$ (a definite integral) are called integrals.

The process of finding either $\,\int f(x)\,dx\,$ or $\,\int_a^b f(x)\,dx\,$ is called integration.

The symbol $\,\int\,$ is called the integral sign.

The function that is being integrated is called the integrand. For example, in the indefinite integral $\,\int 3x^2\,dx\,,$ the integrand is $\,3x^2\,.$

What About that ‘$\,dx\,$’ ?

It may be helpful to view the integral sign ‘$\,\int\,$’ and the ‘$\,dx\,$’ as an inseparable instruction pair. If you've got an integral sign, then you need a $\,dx\,.$ (Of course, other variables can be used.)

The function that is being integrated is placed between the integral sign and the $\,dx\,.$

In finding an indefinite integral, think of the inseparable ‘$\,\int\,dx\,$’ pair as an instruction that is saying:

Find all antiderivatives (with respect to $\,x\,$) of the function that is smushed between us!

A Strategy for Finding Antiderivatives

In general, antidifferentiation is more difficult than differentiation.

The following list of questions/considerations offers a thought process to help!

Is it Just a Differentiation Rule, in Reverse?

The first question you should always consider when analyzing an indefinite integral is:

Do I know a function whose derivative is ‘blah’ ?

Is this just a differentiation formula, ‘in reverse’ ?

Example (A Differentiation Formula, in Reverse)

Do we know a function whose derivative is $\,\sec x\,\tan x\,$?

Yes!  Recall that:

So, $\,\sec x\,$ is an antiderivative of $\,\sec x\, \tan x\,.$

Therefore:

Example (A Differentiation Formula, in Reverse)

Do we know a function whose derivative is $\,\frac{1}{\sqrt{1-x^2}}\,$?

Yes!  Recall that:

So, $\,\arcsin x\,$ is an antiderivative of $\,\frac{1}{\sqrt{1-x^2}}\,.$

Therefore:

Sometimes, You Just Need to Rename

Sometimes, the function being integrated needs to be renamed, using basic algebra skills.

Perhaps you need to write a radical as a rational exponent.

Perhaps you need to use a distributive law.

Perhaps you need to FOIL something out.

Perhaps you need to use the fact that $\frac{A+B}C = \frac{A}{C} + \frac{B}{C}\,.$

Example (Renaming)

Here, we used the fact that, for $\,n\ne -1\,$:

This is often called the Simple Power Rule for Integration.

If a problem starts off in radical form, then you should give your answer in radical form.

Recall that dividing by a fraction is the same as multiplying by its reciprocal. For example, dividing by $\,\frac 23\,$ is the same as multiplying by $\,\frac 32\,.$

Example (Renaming)

In prior differentiation lessons, you learned that $\,\frac{d}{dx}\bigl(\ln x\bigr) = \frac 1x\,.$ Thus, $\,\ln x \,$ is a function with derivative $\,\frac 1x\,.$ What's going on with the appearance of $\ln |x|\,$ in this example, instead of just $\,\ln x\,$?

Here's the problem:  when finding antiderivatives, you want the function being integrated and its antiderivative to have the same domain (or as close as possible).

Unfortunately, the domain of $\,\frac 1x\,$ is all nonzero real numbers, whereas the domain of $\,\ln x\,$ is only all positive real numbers. This undesirable situation is easily remedied, as follows:

The most general antiderivative of $\,\frac 1x\,$ is $\,\ln |x|\,$:

• The derivative of $\ln |x|\,$ is $\,\frac 1x\,$:  It's easy to check that $\,\ln |x|\,$ has derivative $\,\frac1x\,.$ Just write $\,\ln |x|\,$ as a piecewise-defined function, and then differentiate both pieces.
• $\ln |x|\,$ and $\,\frac 1x\,$ have the same domains: both have all nonzero real numbers as their domain.

Substitution

There's a common theme in mathematics:  if you can't solve a problem, try to ‘turn it into’ a problem that you can solve.

Solve the new problem, and then use this answer to solve the original problem. Substitution is just such a method!

Key idea for substitution:  Look for something in the integrand with a derivative that is also a factor in the integrand, perhaps off by a constant.

This idea is illustrated in the examples below. The substitution technique is often called ‘$u$-substitution’ since $\,u\,$ is a common variable used in the renaming process.

The expression chosen for $\,u\,$ is often something in parentheses, something in an exponent, something in a denominator, or something under a radical.

Example (Substitution)

In this integral, note that the derivative of $\,3x^2 - 1\,$ is $\,6x\,.$ The variable part of $\,6x\,$ is $\,x\,.$ There is a factor of $\,\color{red}{x}\,$ in the integrand! Woo hoo!

In other words, the derivative of $\,3x^2 - 1\,$ is also a factor in the integrand; it's just off by a constant. We can deal with things being off by a constant.

Here's what the solution looks like:

Let $\,\color{purple}{u = 3x^2-1}\,,$ so that $\,\frac{du}{dx} = 6x\,.$ Note that $\,\frac{du}{dx}\,$ is Leibniz notation for the derivative of $\,u\,$ with respect to $\,x\,.$

The derivative statement $\,\frac{du}{dx} = 6x\,$ ‘reformulates’ to $\color{red}{\,du = 6x\ dx\,}\,.$

With $\,\color{purple}{u = 3x^2-1}\,$ and $\color{red}{\,du = 6x\ dx\,}\,$ we have:

Note how we multiply by $\,1\,$ in the form of $\,\frac{6}{6} = 6\cdot\frac{1}{6}\,.$ The $\,6\,$ is left inside the integral (where it becomes part of $\,du\,$) but the $\,\frac{1}{6}\,$ is slid out of the integral. Remember: only constants can be slid out of integrals—not anything involving the variable of integration!

Note also that we started with an integral in $\,x\,,$ so we must end up with the answer in $\,x\,.$

Example (Substitution)

Recall that $\,\sin^5(3x)\,$ is a shorthand for $\,\bigl(\sin(3x)\bigr)^5\,.$

The derivative of $\,\sin(3x)\,$ is $\,3\cos(3x)\,.$ The variable part of the derivative is $\,\color{red}{\cos(3x)}\,,$ which is a factor in the integrand.

When using substitution, most people skip the ‘$\,\frac{du}{dx} = \text{stuff}\,$’ part and jump right to ‘$\,du = (\text{stuff})\,dx\,$’ .

Here's the solution:

Let $\,u = \sin(3x)\,$, so that $\color{red}{\,du = 3\cos(3x)\,dx\,}$. Then:

Integration by Parts

Integration by Parts is the integration counterpart to the Product Rule for Differentiation, as follows:

Let $\,u\,$ and $\,v\,$ both be functions of $\,x\,,$ so that:

Integrating both sides with respect to $\,x\,$ gives:

This simplifies to:

A bit of rearrangement gives the Integration by Parts Formula:

Observations About the Integration By Parts Formula

Here are some observations regarding the Integration by Parts formula:

If Other Methods Fail...

Can't solve an integral by prior methods? Then, try Integration by Parts.

The variables $\,u\,$ and $\,v\,$ look a lot alike. I thought about using different variables, but that would really be going against convention and tradition. So, be sure to look carefully!

Hopefully the New Integral is Easier

When you use the Integration by Parts formula, the hope is that the new integral $\int v\,du\,$ is easier to work with than the original integral $\,\int u\,dv\,.$

You Must Choose $\,u\,$ and $\,dv$

To use the Integration by Parts formula, you must rename the original integral in the form $\,\int u\,dv\,.$ That is, you must choose $\,u\,$ and $\,dv\,.$ Then:

• $u\,$ is differentiated to find $\,du$
• $dv\,$ is integrated to find $\,v$

  Usually (but not always), the constant of integration (in going from $\,dv\,$ to $\,v\,$) is chosen to be zero.

Choose Something for $\,dv\,$ That You Know How to Integrate

First, choose some part of the original integral to be $\,dv\,.$ (The ‘$\,dx\,$’ from the original integral is always part of $\,dv\,.$)

You must choose something for $\,dv\,$ that you know how to integrate. Why? If you can't integrate $\,dv\,$ to find $\,v\,,$ then you can't use the right side of the Integration by Parts formula:

Once $\,dv\,$ has been chosen, everything else in the integrand becomes $\,u\,.$

What if There are Several Choices for $\,dv\,$?

Is there more than one choice for $\,dv\,$? That is, is there more than one part of the original integrand that you know how to integrate?

If so, then go onto a second consideration—choose something for $\,u\,$ that gets simpler when you differentiate it.

Everything should become clearer with a couple examples:

Example (the Classic Integral that Requires Integration by Parts)

The classic integral that requires Integration by Parts is integrating the natural logarithm function.

Here, there are only two possible choices for $\,dv\,$:

1. $dv = dx$
2. $dv = \ln x\ dx$

If we choose (2), then we're back where we started—unable to integrate the natural log! So, we must choose (1). Everything other than $\,dx\,$ then becomes $\,u\,,$ so $\,u = \ln x\,.$

The solution looks like this:

Let $\,dv = dx\,,$ so that $\,v = x\,.$

Then $u = \ln x\,,$ so that $\,du = \frac 1x\,dx\,.$

Use the Integration by Parts formula:

Let's verify that $\,x\ln x - x\,$ is indeed an antiderivative of $\,\ln x\,$:

Example (Integration by Parts)

Here, there are actually four possible choices for $\,dv\,$:

1. $dv = dx$   (here, $\,u\,$ would be $\,x{\text{e}}^x\,$)
2. $dv = x\,dx$   (here, $\,u\,$ would be $\,{\text{e}}^x\,$)
3. $dv = {\text{e}}^x\,dx$   (here, $\,u\,$ would be $\,x\,$)
4. $dv = x{\text{e}}^x\,dx$   (here, $\,u\,$ would be $\,1\,$)

Discard (4). Of the remaining three, which leaves us with a ‘$\,u\,$’ that gets simpler when you differentiate it? Only (3)!

So, here's the solution:

Let $u = x\,,$ so that $\,du = dx\,.$

Then $\,dv = {\text{e}}^x\,dx\,,$ so that $\,v = {\text{e}}^x\,.$

Then:

In this exercise, all functions are assumed to have the required properties for a particular situation.

For example, any function in a denominator is assumed to be nonzero, any function inside a logarithm is assumed to be positive, and so on.