MATH 155 SUPPLEMENTAL NOTES 12

BASIC INTEGRATION FORMULAS

This set of supplemental notes will review some of the basic integration formulas, or as I call them, "tricks" that we can use when integrating certain functions. There are seven basic tricks and they are the following:

u-substitution
Completing the square
Using a trigonometric identity
Eliminating a square root
Reducing an improper fraction
Separating a fraction
Multiplying by a form of 1

In these notes, I will provide several examples of how to use each trick.

U-SUBSTITUTION

EXAMPLE 1:

SOLUTION:

Let u = 1 + 3sin x, then du = 3cos x dx.

EXAMPLE 2:

SOLUTION:

EXAMPLE 3:

SOLUTION:

Before we try to determine what u is going to be, let us simplify the given function.

Now the integral becomes the following.

Let u = e ^x, then du = e ^xdx. Now notice that this integral is of the form u ² + a ², which the form for the inverse tangent integral.

EXAMPLE 4:

SOLUTION:


	Let a = 1 and p = 2u, then dp = 2du. We can also determine from this that the integral is of the form a ² - p ², which is the pattern for the inverse sine integral.

EXAMPLE 5:

SOLUTION:

First of all, to be able to do a u-substitution, I am going to multiply both top and bottom

by a factor of e ^y.

Let u = e ^y, then du = e ^ydy. You should notice that this pattern tells us that we are looking at the inverse secant integral.

COMPLETING THE SQUARE

Sometimes we will need to complete the square of a quadratic equation, and then we will look at the result to see what inverse trig function pattern it fits.

EXAMPLE 6:

SOLUTION:

First of all, I am going to complete the square.

x ² - 6x + 10 ® (x ² - 6x ) + 10 ® (x ² - 6x + 9) + 10 - 9 ® (x - 3) ² + 1

So now substitute this in for x ² - 6x + 10.

Let a = 1 and u = x - 3, then du = dx. When x = 2, then u = -1, and when x = 4, then u = 1. Notice that the denominator is of the form u ² + 1, therefore this is the inverse tangent function.

EXAMPLE 7:

SOLUTION:

First of all, I will complete the square.

2x - x ² ® - (x ² - 2x ) ® 1 - (x ² - 2x + 1) ® 1 - (x - 1) ²

Now I will replace 2x - x ² with 1 - (x - 1) ².

Let a = 1 and u = x -1, then du = dx. Notice that this pattern is for the inverse sine function, so use the formula for that integral.

EXAMPLE 8:

SOLUTION:

Again, let us complete the square.

x ² - 4x + 3 ® (x ² - 4x ) + 3 ® (x ² - 4x + 4) + 3 - 4 ® (x - 2) ² - 1

Now to replace x ² - 4x + 3 with (x - 2) ² - 1.

Let a = 1 and u = x - 2, then du = dx and this integral fits the pattern for the inverse secant function.

USING A TRIGONOMETRIC IDENTITY

EXAMPLE 9:

SOLUTION:

First of all, we cannot integrate sin ²x as is, but we can use a trig identity to simplify this integral.

FACT: sin ²x = (1/2)(1 - cos 2x)

FACT: cos ²x = (1/2)(1 + cos 2x)

So I am going to replace sin ²x with its related identity.

EXAMPLE 10:

SOLUTION:

I am going to substitute 2sin x cos x in for sin 2x.

FACT: sin 2x = 2sin x cos x

Let u = sin x, then du = cos x dx.

I could have integrated the original integral directly, and the answer would look like the following.

Which of the two answers is correct? They both are correct. The only difference between them is that they differ by a constant.

EXAMPLE 11:

SOLUTION:

Well, the problem with this integral is that sin 3x does not have a direct substitution. But there is identity for the sin (a + b).

FACT: sin (a + b) = sin a cos b + cos a sin b

So sin 3x can be thought of as sin (2x + x).

Now to substitute csc x sin 3x with 1 + 2cos 2x.

ELIMINATING A SQUARE ROOT

Most of these problems involve both a square root and trig functions, so to eliminate the square root, we may have to use some trig identities.

EXAMPLE 12:

SOLUTION:

Recall that

Therefore, I will substitute 2sin ²x in for 1 - cos 2x.

EXAMPLE 13:

SOLUTION:

Recall that sin ²x + cos ²x = 1, then cos ²x = 1 - sin ²x. Therefore, I will replace 1 - sin ²x with cos ²x.

REDUCING AN IMPROPER FRACTION

Recall that an improper fraction is a fraction where the degree of the numerator is greater than or equal to the degree of the denominator. To reduce an improper fraction you must use long division.

EXAMPLE 14:

SOLUTION:

First thing that I need to do with the integrand is to reduce the improper fraction by performing long division.

EXAMPLE 15:

SOLUTION:

Again, I will perform long division to simplify this improper fraction.

Let u = 2x + 3, then du = 2 dx. When x = -1, then u = 1, and when x = 3, then u = 9.

SEPARATING A FRACTION

We will use this idea in simplifying fractional integrands.

EXAMPLE 16:

SOLUTION:

Let u = x - 1, then du = dx.

EXAMPLE 17:

SOLUTION:

Let u = 2x, then du = 2dx. When x = 0, then u = 0, and when x = 1/2 then u = 1.

Let p = 1 + 4x ², then dp = 8x dx. When x = 0, then p = 1, and when x = 1/2, then p = 2.

Notice that the first integral of the above line is of the form of the inverse tangent function.

MULTIPLYING BY A FORM OF 1

There are a few little tricks that we do in algebra. One is adding or subtracting a form of zero to an equation, and the other is multiplying by a form of 1. The most common use of multiplying by a form of 1 is when we are rationalizing the denominator of a radical, rational expression. We will now use this method to simplify the integrand of an integral.

EXAMPLE 18:

SOLUTION:

Why did I multiply both top and bottom by sec x + tan x? Well, suppose that y = sec x + tan x, then y' = sec x tan x + sec ²x. So, if I let u = sec x + tan x, then I will have the value for du.

Let u = sec x + tan x, then du = sec ²x + sec x tan x.

FACT:
FACT:

EXAMPLE 19:

SOLUTION:

First of all, I am going to multiply both top and bottom by the conjugate of 1 + cos x.

EXAMPLE 20:

SOLUTION:

Again, let us multiply the top and bottom of this fraction by the conjugate.

Recall that cot ²x + 1 = csc ²x, so csc ² x - cot ² x = 1.

Work through these examples, making note of the "tricks" that were used to simplify the integrand. As always, the more you work with these "tricks", the easier it will be to figure out which one you will have to use. If you have any questions on any of this problems, feel free to contact me.

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