Derivative and Integral RulesThese are on the inside of the back cover of your text.

General Derivative Rule General Integral Rule

u(x)r = r u(x)r - 1 u�(x)d

dx � u(x)r u�(x) dx = + Cu(x)r�1

r�1

r � -1

eu(x) = eu(x) u�(x)d

dx � eu(x) u�(x) dx = eu(x) + C

bu(x) = ln(b) bu(x) u�(x)d

dx� bu (x ) u �(x ) dx � bu (x )

ln(b ) � C

logb(u(x)) =d dx

u�(x)ln(b) u(x)

u � (x )u(x )

dx � ln(b) logb�u (x )� � C

ln(u(x)) = d dx

u�(x)u(x)

� dx = ln(u(x)) + Cu�(x)u(x)

� dx = ln(�u(x)�) + Cu�(x)u(x)

General Derivative Rule General Integral Rule

sin(u(x)) = cos(u(x)) u�(x)d dx � cos(u(x)) u�(x) dx = sin(u(x)) + C

cos(u(x)) = - sin(u(x)) u�(x)d dx � sin(u(x)) u�(x) dx = - cos(u(x)) + C

tan(u(x)) = sec2(u(x)) u�(x) d dx � sec2(u(x)) u�(x) dx = tan(u(x)) + C

cot(u(x)) = - csc2(u(x)) u�(x) d dx � csc2(u(x)) u�(x) dx = - cot(u(x)) + C

sec(u(x)) = sec(u(x)) d dx

tan(u(x)) u�(x)

� sec(u(x)) tan(u(x)) u�(x) dx =sec(u(x)) + C

csc(u(x))d dx

= - csc(u(x))cot(u(x)) u�(x)

� csc(u(x)) cot(u(x)) u�(x) dx= - csc(u(x)) + C

Which of the following can you integrate using theChain Rule and introducing a multiplicative constant?

� g (x) dx � 1k�

k � g (x) dx

The object is to express the integral in a form that fitsone of the general rules:

� g (x) dx � 1k�

u�(x) � f (u(x)) dx

a �e 5xdx

b �2sin(3t) dt

c �r �1r 3

dr

d �1

t 1/2 dt

e �t 2�1

t 3�3t�5

dt

f �cos(4s) ds

(a) �e �3 tdt

(b) �2 sin(3 t) dt

(c) �6t 3

dt

(d) � 5

t dt

(e) �

6t 2 � t

dt

(f) �cos(sin( t)) dt

Which can you integrate?What Rule do you use?

�6x � 2

x 3 dx

�( t�3) t 2 � 6t � 1 dt

�6 cos( t)

sin(t) dt

Algebraic Manipulation: Trig identities

�tan2 ( t) dt

�sin3(t) dt

a �sin(t) e �3cos(t)dt

b �5e 3 t sin(e 3 t) dt

c �4 ln(6)ln(5)

t 3 dt

d � 5 t 2/3 dt

e �

6t�3(t 2 � t) 5

dt

f �cos(t) cos(sin( t)) dt

Chain Rule– Substitution

Introduction. Evaluating an integral that is not in the form of abasic general integral formula.

First approach. Is it possible to manipulate theintegrand to obtain an equivalent integral that is of aform that can be directly evaluated.

We consider three types of integrals.1. I = � f (x) dx

When f(x) has a distinguished sub-term asubstitution method may work. If the sub-term u(x)and u�(x) is present, when

a. u = a + b xr substitute u

b. u = eh(x) substitute u

c. u = ±a2 ± x2 Use trig-substitutions.

2. For I = � f (x)� g(x) dxWhen g(x) � f�(x) the integral can be transformed

by the method of Integration of Parts (IP), based onthe product rule

(u�v)� = u��v + u�v�

Solving for one term on the right gives

u��v = _______ _________

so, upon integrating,

�u��v dx = �_____dx - �_____dx

The FTC gives �(u�v)�dx = _____ .So, the IP formula gives the integral of u��v as theterm u�v minus the integral of u�v�.IP:

3. When the integral involves a fraction.

The integral I = f (x)g(x)

dx

can be evaluated directly only if

f(x) = ______________

or, if

g(x) = ur(x) and f(x) = ____________

Otherwise, one must reduce the fraction to obtain oneof these cases to evaluate I.

The basic principle is to write a fraction as the sumof two simpler (partial) fractions:

5/6 = 1/2 + 1/3

§6.1 Substitutions. Theory.

Applying the general integration rules involvesmaking a simple substitution.

To integrate an integral

I = � f (x) dxusing one of the General Integration Rules you mustrecognize a term u(x) and a function F (x) so that theintegral has the form

= _______________�F �(u) u �(x) dx

The integral form of the Chain Rule.

If the choice of the function F is not obvious theintegral may be evaluated by first manipulating theintegrand into a different form.

If there is a distinguished sub-term u(x) one can try a substitution u = u(x).

To transform the integral with respect to x to anintegral with respect to u we need to express f(x) anddx in terms of u and du.

If u = u(x) that can be solved for x = g(u).

Then dx = ________and

I = � f (g(u)) g�(u) du

The function u is chosen so that f(g(u)) g�(u)becomes a nice/simple function of u.

This method works only if the resulting u-integralcan be easily evaluated using “known” integralformulas.

Then, the antiderivative must be transformed back toa function of x by setting u = g(x).

§6.1 Substitutions. Examples.

Linear substitution. u = ax + b

Evaluate I = � x( 2 - x)1/3 dx

Exponential Substitution. u = eh(x).

Evaluate I = � (e2x - 3)2 e2x dx

Trigonometric Substitutions for

a2 + x2 , a2 - x2, or x2 - a2.Integrals involving quadratic functions can often

be evaluated by using simple "trigonometric"

substitutions.

These substitutions "work" is because of the

Pythagorean Identities:

sin2(θ) + cos2(θ) = 1

tan2(θ) + 1 = sec2(θ)

or sec2(θ) - 1 = tan2(θ)

There are three basic trig-substitutions, depending on

the form of the integrand.

SINE SUBSTITUTION: For a2 - x2

use the substitution: x = a sin(θ)

and dx = _________

Then,

a2 - x2 = ________________

= _________________

= ___________.

Then

θ = sin-1( ) or θ = Arcsin( )xa

xa

TANGENT SUBSTITUTION:

For a2 + x2

The substitution is: x = a tan(θ)

and dx = a sec2(θ) dθ,

Then,

a2 + x2 = _________________

= __________________

= _______________.

in this case θ = tan-1( )xa

or θ = Arctan( )xa

SECANT SUBSTITUTION: For x2 - a2

The substitution: x = a sec(θ)

and dx = a sec(θ)tan(θ) dθ

Then,

x2 - a2 = ____________________

= ___________________

= _______________.

θ = sec-1( ) or θ = Arcsec( )xa

xa

0-0-0-

a a2

- x2

a

x2

- a2

xx

a2

+ x2

a x

x = a tan(0) x = a sin(0) x = a sec(0)

The three trig-substitutions result in integrals with

respect to θ.

After integrating the antiderivtive must be converted

back from the θ-variable to the x-variable.

This can be done two ways:

i)

or

ii) using right triangles having angle θ and side

lengths x and constant a:

Example. Evaluate 1

1�9x 2 dx

Evaluate I = 4x�1(x�4)2

dx

§6.2 Integration by Parts (IP) Theory.

When an integral involves a product of terms but is

not of the form

�F �(u(x)) u �(x) dx

the integral usually can not be

evaluated by inspection.

An integral of the form

I = � f (x)� g(x) dx

May be evaluated using the method of integration by

parts.

Integration by Parts, or IP for short, consists of

identify the factors f(x) and g(x) as a function u(x)

and a derivative v�(x). When this is done the integral

is evaluated as

I = �u�v� dx = �(u�v)�dx - �u��v dx

or

I = �u�v� dx = u�v - �u��v dx

What does the IP formula do?

When will it “work”?

When will it “not work”?

When you apply the IP formula you must make a

choice.

Which term is u(x) and which is v�(x)?

I suggest you make a templet to help organize your

choice and work.

u = ____________ v� = ____________

u� = _ _ _ _ _ _ _ _ v = _ _ _ _ _ _ _ _ _

Then set

I = �u�v� dx = u�v - �u��v dx

and evaluate the integral �u��v dx.

How do you choose u and v�?

Requirement 1. You must be able to integrate v�.

Requirement 2. The derivative of u should not be

more complex.

Requirement 3. You should be able to evaluate

�u�v� dx.

What do you do if you can not evaluate the resulting

integral �u��v dx ?

Option 1. Consider the reverse choice of u and v�.

Option 2. Press forward. Try it again!!

Careful: repeated use of the IP method

could take you back to the original

problem. If this happens reverse your

choice for u and v�

§6.2 Integration by Parts (IP) Examples.

Evaluate I = � 3x e-x dx

u = ____________ v� = ____________

u� = _ _ _ _ _ _ _ _ v = _ _ _ _ _ _ _ _ _

Evaluate I = � 3x ln(x) dx

u = ____________ v� = ____________

u� = _ _ _ _ _ _ _ _ v = _ _ _ _ _ _ _ _ _

Evaluate I = � sin2(2x) dx

u = ____________ v� = ____________

u� = _ _ _ _ _ _ _ _ v = _ _ _ _ _ _ _ _ _

Evaluate I = � sin(2x) e3x dx

u = ____________ v� = ____________

u� = _ _ _ _ _ _ _ _ v = _ _ _ _ _ _ _ _ _

Evaluate I = � 3x2 e-x dx

u = ____________ v� = ____________

u� = _ _ _ _ _ _ _ _ v = _ _ _ _ _ _ _ _ _

§6.3 Partial Fractions. Theory

Only two general integral formulas involve rational

expressions, i.e., fractions.

The obvious formula is

A less obvious formula is

To evaluate

I = f (x)g(x)

dx

we must manipulate it to obtain integrals of one of

the two formulae types.

Four manipulations are frequently useful.

1. Cancellation: If f(x) and g(x) have common

factors, cancel them to reduce

the complexity.

2. Reduction of the fraction:

Division: MN

� A � RN

when M = A�N + R

3. Separate the numerator: A � BC

� AC

� BC

4. Partial Fractions: Write the fraction f/g as the sum

of two fractions:

This is possible only if the denominator g(x) is the

product of two terms, say Q1 and Q2.

Then, you can write

fg

� fQ1�Q2

� P1

Q1

�P2

Q2

If g(x) = Q1(x)�Q2(x) how do you express the fraction

f/g as a partial fraction?

You use the factors Q1 and Q2 as the denominators

and find the numerators P1 and P2 algebraically using

the fact that

fQ1�Q2

� P1

Q1

�P2

Q2

� Q1�Q2

if and only if

f(x) =____________________

for all values of x.

A rational function that is a fraction f/g of two

polynomials is said to be in reduced form if

degree( f ) < degree( g )

If a rational expression is not in reduced form,

dividing the denominator into the numerator will

give a polynomial plus a fraction that is in reduced

form:

f(x)g(x)

� P(x) �R(x)g(x)

where f(x) = P(x)�g(x) + R(x)

and degree( R(x) ) < degree( g(x) )

Equating Polynomials

Two polynomials are equal if, and only if, their

coefficients of each power of x are identical.

= �� ai = bi for all i.�N

i�0ai x

i �N

i�0bi x

i

What is the form of P(x) if

Q(x) is a polynomial?

Assuming that f/g is of reduced form, and Q(x) is a

simple factor of g(x) then the partial expansion of f/g

will have a factor

P/Q where P(x) is a polynomial of degree less than

the degree of Q(x).

E.g., if a factor is Q(x) = 2x2 + 1 then the partial

expansion will contain a term of the form

Ax�B2x 2

�1

If a factor is linear, say Q(x) = x + 5 then the partial

expansion will contain a term of the form Ax�5

Repeated Factors

If the denominator has a factor raised to a power this

factor is called a repeated factor.

The power is called the factor’s multiplicity.

When expanding a fraction with a repeated fraction

you must include a term with each power of the

repeated factor up to its multiplicity.

For example, if a factor of g(x) is

Q(x) = (x - 1)3

i.e., the factor (x - 1) with multiplicity 3, then the

partial expansion will contain the terms

+ + Ax�1

B(x�1)2

C(x�1)3

§6.3 Partial Fractions. Examples.

Ask these questions:

1. Is cancellation possible?

2. Is it of reduced form?

3. What are the factors of the denominator?

4. What is the form of the partial fraction

expansion?

Example.

Evaluate I = 2x 2�x

(2x�1)(x 2�4)

dx

Example Evaluate I = 3x 2�2x�5x 3

dx

Example Evaluate I = x 3

3x 2�2x�5

dx

Evaluate I = (2x�1)x 3

(2x�1)(x 2�4)

dx

Evaluate I = 4x�1(x�4)2

dx