SUNY Geneseo Department of Mathematics

Integration by Substitution, Part 1

Monday, November 23

Math 221 02
Fall 2020
Prof. Doug Baldwin

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Anything You Want to Talk About?

See below.

SI

The next session is tonight at 6:00, featuring open-ended discussion of whatever you want to talk about.

Integration in Mathematica

Use the Integrate function to calculate both definite integrals and indefinite integrals (antiderivatives).

The syntax for an indefinite integral (antiderivative) is

Integrate[ expression, variable ]

where expression is the expression or function to integrate, and variable is the variable in it. The syntax for a definite integral is

Integrate[ expression, { variable, low, high } ]

where expression and variable are the same as for an indefinite integral, and low and high are the bounds of the interval to integrate over.

This notebook contains examples of both kinds of integral.

Finding Antiderivatives by Substitution

From the beginning of section 5.5 in the textbook and the substitution discussion.

Examples

One way to understand substitution is to think of it as simplifying antiderivatives (also known as indefinite integrals) by spotting places where awkward expressions and their derivatives can temporarily be replaced by variables.

For example, imagine trying to integrate (x-4)¹⁰. This looks difficult (maybe you have to multiply out (x-4) to the tenth power), but if the x-4 part were only a single variable, it would be easy with the power rule for antiderivatives:

Substitution simplifies an integral, but leaves 2 variables in it

The fact that we’re taking an antiderivative involving u but integrating with respect to variable x is a sign that something is wrong: antiderivatives need the “dx” part (the differential) to agree with the variable in the integrand. But this is fixable, if we think about du and dx as differentials (i.e., separate quantities that can be moved around equations independently of each other), and express dx in terms of du:

Calculating du in terms of dx rewrites integral solely in terms of u

Now that we’ve discovered that du and dx are in fact equal, we can replaced dx in the integral with du, and the rest is just using the power rule for antiderivatives. And at the end, replace u with the corresponding expression involving x, since we started with a function of x:

Integrate in terms of u and then substitute back

Sometimes the integrand contains a multiple of du rather than exactly du. For example, consider this integral:

Integral of x squared times the quantity 1 plus 2 x cubed all cubed

Letting u represent 1 + 2x³ looks promising in terms of getting to a place where we could use the power rule again. But its derivative is 6x² dx, and the integrand only contains x² dx. But since x² dx is 1 sixth of du, replace it with that, literally:

Integral becomes u cubed times 1 sixth d u

Now integrate in terms of u, replacing u with 1 + 2x³ at the end:

Integrate 1 sixth u cubed then replace u with 1 plus 2 x cubed

This idea of manipulating expressions involving u and x in order to put an integrand entirely in terms of u can be taken further. For example, consider this integral, using the substitution u = x - 2 and thus du = dx:

Integral of x times square root of x minus 2

Applying the substitution to the integrand still leaves an x present, and we can’t integrate until everything is in terms of u. But thinking back to the substitution, if u = x - 2, then x = u + 2:

Replace x with u plus 2

Now we can simplify the integrand with algebra, and then integrate using the power rule:

Integrate u plus 2 all times square root of u

Examples Solved in the Discussion

The following 2 examples had good solutions in the Canvas substitution discussion. Here they are with a few details written out in a style similar to the above examples:

Integrate x times cosine x squared by the substitution u equals x squared

Integrate 2 x over x squared plus 4 by the substitution u equals x squared plus 4

Some Trickier Examples

Integrate 2 sinx cosx.

There are many different ways you could do this. First, try letting u = sin x:

Integrate 2 times sin times cosine via substitution u equals sine x

But you could just as well let u = cos x:

Integrate 2 sine x times cosine x via substitution u equals cosine x

This looks bad, because we seem to have gotten a different answer from the other substitution. But thanks to a trigonometric identity, it turns out that -cos²x and sin²x just differ from each other by a constant, and that’s allowed by the constant of integration C (it really does turn out to matter!)

Sine squared plus cosine squared equals 1 means minus cosine squared equals sine squared minus 1

You could even evaluate this integral by using the trigonometric identity 2 sinx cosx = sin(2x), which leads to a simpler substitution, but more work with trigonometric identities both before and after:

Integrate 2 sine x times cosine x as sine of 2 x then substitution u equals 2 x

Keep practicing substitution via the substitution discussion. There’s no new reading, but do practice substitution via the discussion.

Remember that our next meeting is the optional Zoom “class” on Monday, November 30. It will provide a further opportunity to talk about or answer questions about the substitution discussion, after which I’ll post the results of that talking and answering to Canvas. I’ll try to invite you all to the Zoom via Google calendar and email, so watch your email for joining information.

Have a good break!

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