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Use pattern recognition to find an indefinite integral.

Use a change of variables to find an indefinite integral.

Use the General Power Rule for Integration to find an indefinite integral.

Objectives

ANTIDIFFERENTIATION OF A COMPOSITE FUNCTIONLet g be a function whose range is an interval I and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then

If u = g(x), then du = g’(x)dx and

( ( )) '( ) ( ( ))f g x g x dx F g x c

( ) ( )f u du F u c

PATTERN RECOGNITION

In this section you will study techniques for integrating composite functions.

This is split into two parts, pattern recognition and change of variables.

u-substitution is similar to the techniques used for the chain rule in differentiation.

PATTERN RECOGNITION FOR FINDING THE ANTIDERIVATIVEFindLet g(x)=5x and we have g’(x)=5dx So we have f(g(x))=f(5x)=cos5xFrom this, you can recognize that the integrand follows the f(g(x))g’(x) pattern. Using the trig integration rule, we get

You can check this by differentiating the answer to obtain the original integrand.

5cos5xdx

cos5 5 sin 5x dx x c

RECOGNIZING PATTERNS

2 4 22 ( 1) 1 2x x dx u x du x

Look at the more complex part of the function- the stuff ‘inside’.

Does the stuff outside look like the derivative of the stuff inside?

52 1

5

xc

RECOGNIZING PATTERNS

2 3 3 23 1 1 3x x dx u x du x

3/ 2321

3x c

RECOGNIZING PATTERNS

2 2sec tan 3 tan 3 secx x dx u x du x

2tan 3

2

xc

CHANGE OF VARIABLES

FindLet u=5x and we have du=5dx

You can check this by differentiating the answer to obtain the original integrand.

5cos5xdx

cos5 5 sin 5x dx x c

cos sinudu u c

CHANGE OF VARIABLESA.K.A. U-SUBSTITUTIONExample 3 222 1x x dx

2 1 2u x du xdx 222 1x x dx

2u du 2 1 3

2 1 3

u uc c

32 1

3

xc

PATTERN RECOGNITION

The integrands in Example 1 fit the f(g(x))g'(x) pattern exactly—you only had to recognize the pattern.

You can extend this technique considerably with the Constant Multiple Rule

Many integrands contain the essential part (the variable part) of g'(x) but are missing a constant multiple.

In such cases, you can multiply and divide by the necessary constant multiple, as shown in Example 3.

MULTIPLYING AND DIVIDING BY A CONSTANTExample 3 22 1x x dx

2 1 2u x du xdx 221

2 12

x x dx

2 31 1

2 6u du u c

321now back substitute : 1

6x c

MULTIPLYING AND DIVIDING BY A CONSTANT ALTERNATE METHODExample 3 22 1x x dx

2let 1 2u x du xdx but we have only , not 2 ...so we do algebrax x

1now substitute :

2du xdx

2 1

2u du

2 31 1

2 6u du u c

321now back substitute : 1

6x c

CHANGE OF VARIABLESEXAMPLE 4

2 1x dx

EXAMPLE 4 – CHANGE OF VARIABLES

Find

Solution:

First, let u be the inner function, u = 2x – 1.

Then calculate the differential du to be du = 2dx.

Now, using

and

substitute to obtain

EXAMPLE 4 – SOLUTIONcont’d

CHANGE OF VARIABLES

Example 5 2 1x x dx

CHANGE OF VARIABLES

Example 5 2 1x x dx_ 2 1 2let u x du dx

: ( 1) / 2solve for x u 1

21

: 2 12 2

u duSubstitute x x dx u

1 3 1

5/ 2 3/ 22 2 21 1 1 2 2

14 4 4 5 3

u u du u u du u u c

5 3

2 21 1

_ : 2 1 2 110 6

Back Substitute x x c

CHANGE OF VARIABLES EXAMPLE

Example 6

2sin 3 cos3x xdx

EXAMPLE 7 – SUBSTITUTION AND THE GENERAL POWER RULE

EXAMPLE 7 – SUBSTITUTION AND THE GENERAL POWER RULE cont’d

GUIDELINES FOR MAKING A CHANGE OF VARIABLES1. Choose a substitution u=g(x). Usually it is

best to choose the INNER part of a composite function, such as a quantity raised to a power.

2. Compute du=g’(x)dx

3. Rewrite the integral in terms of the variable u

4. Find the resulting integral in terms of u

5. Replace u by g(x) to obtain an antiderivative in terms of x.

6. Check your answer by differentiating.

GENERAL POWER RULE FOR INTEGRATIONTheorem: The General Power Rule for Integration

If g is a differentiable function of x, then

1

Equivalently, if u=g(x), then

1

,1

nn g x

g x g x dx c nn

1

, 11

nn uu du c n

n

CHANGE OF VARIABLES FOR DEFINITE INTEGRALS

If the function u=g(x) has a continuous derivative on the closed interval [a,b] and f is continuous on the range of g, then

Basically, this gives us permission to find the area under the curve with these types of integration problems.

b g b

a g af g x g x dx f u du

DEFINITE INTEGRALS AND CHANGE OF VARIABLES Example 8

1 32

01x x dx

EXAMPLE 8 – CHANGE OF VARIABLESEvaluate

Solution:

To evaluate this integral, let u = x2 + 1.

Then, you obtain

Before substituting, determine the new upper and lower limits of integration.

EXAMPLE 8 – SOLUTION

Now, you can substitute to obtain

cont’d

EXAMPLE 8 – SOLUTION

Try rewriting the antiderivative

in terms of the variable x and evaluate the definite integral at the original limits of integration, as shown.

Notice that you obtain the same result.

cont’d

DEFINITE INTEGRALS AND CHANGE OF VARIABLESExample 9:

This problem requires a little more work: 5

1 2 1

xdx

x

INTEGRATION OF EVEN AND ODD FUNCTIONSEven with a change of variables, integration can be difficult.

Occasionally, you can simplify the evaluation of a definite integral over an interval that is symmetric about the y-axis or about the origin by recognizing the integrand to be an even or odd function (see Figure 4.40).

Figure 4.40

INTEGRATION OF EVEN AND ODD FUNCTIONS

EXAMPLE 10 – INTEGRATION OF AN ODD FUNCTION

Evaluate

Solution:

Letting f(x) = sin3x cos x + sin x cos x produces

f(–x) = sin3(–x) cos (–x) + sin (–x) cos (–x)

= –sin3x cos x – sin x cos x = –f(x).

EXAMPLE 10 – SOLUTION

So, f is an odd function, and because f is symmetric about the origin over [–π/2, π/2], you can apply Theorem 4.16 to conclude that

Figure 4.41

cont’d

HOMEWORK:

p.304 1,3,9,14,16,20,25,27,31,33,34,41,45,51,54,59,69,79,81,95,103

These are the minimum to help you PASS this next quiz/test. I suggest you complete all problems in the sections relating to what was covered in the notes.

Students taking the AP Exam in the spring should also look at 107, 108, 111-112, and the word problems (see me for assistance when you get to them).