周次: 5 日期: March 25, 2022 节次: 2

In this course, we will discuss the definition of hyperbolic functions. And we shift our attention to several integration techniques. These techniques can help us integrate more functions. A typical technique called “Integration by Parts” will be discussed as well.

§ 7.8 A Brief Introduction on Hyperbolic Functions

Let’s recall the odd and even functions. We have stated that for every function f defined on an interval centered at the origin can be written in a unique way as the sum of one even function and one odd function. The decomposition is

$$ f(x) = \underbrace{\frac{f(x) + f(-x)}{2}}{\text{even part}} + \underbrace{\frac{f(x)-f(-x)}{2}}{\text{odd part}} $$

No exception for exponential function ex , it can be decomposed. And we assigned two new functions to the decomposition.

cosh(x):=even part of ex=ex+ex2

sinh(x):=odd part of ex=exex2

cosh(x) is pronounced as “kosh x”, rhyming with “gosh x”, and sinh(x) is pronounced as “cinch x”, rhyming with “pinch x”.

Graphs of Hyperbolic Functions

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Identities for Hyperbolic Functions

cosh2(x)sinh2(x)=1

sinh(2x)=2sinh(x)cosh(x)

cosh(2x)=cosh2(x)+sinh2(x)

Other Hyperbolic Functions

Analogous to trigonometric functions, the tanh(x), coth(x), sech(x) and csch(x) can all be defined, following the same pattern.

tanh(x):=sinh(x)cosh(x),coth(x):=cosh(x)sinh(x),etc.

The Derivative and Anti-derivative of Hyperbolic Functions

ddxsinh(x)=cosh(x),ddxcosh(x)=sinh(x)

Therefore the anti-derivatives read

sinh(x)=cosh(x)+C,cosh(x)=sinh(x)+C.

Chapter 8. Techniques of Integration

In this chapter we study a number of important techniques for finding indefinite integrals of more complicated functions

§ 8.1 Basic Integration Formulas

Try to match integrals that confront us against one of the standard types. This usually involves a certain amount of algebraic manipulation as well as use of the substitution rules.

Recall that the substitution rule is

f(g(x))g(x)dx=f(u)du,

where u=g(x) is a differentiable function whose range is an interval I and f is continuous on I. Success in integration often hinges on the ability to spot what part of the integrand should be called u in order that one will also have du, so that a known formula can be applied. This means that the first requirement for skill in integration is a thorough mastery of the formulas for differentiation.

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**Note** Students should have a good mastery of formulas 12, 13, 18, 19. For 20 and 22, we provide an alternative (equivalent) anti-derivative

duu2±a2=ln|u+u2±a2|+C

For 20, we have an alternative way to show the anti-derivative, by applying the integration table.

duuu2a2=duu21(au)2=1ad(au)1(au)2=1aarcsin(au)+C

Example 1. Making a Simplifying Substitution

Evaluate

2x9x29x+1dx

Example 2. Completing the Square

Evaluate

dx8xx2dx

Example 3. Expanding a Power and Using a Trigonometric Identity

Evaluate

(secx+tanx)2dx.

Example 4. Eliminating a Square Root

Evaluate

0π/41+cos4x,dx.

Example 5. Reducing an Improper Fraction

Evaluate

3x27x3x+2,dx.

Example 6. Separating a Fraction

Evaluate

3x+21x2,dx.

Example 7. Integral of y=sec(x) — Multiplying by a Form of 1

Evaluate

secx,,dx

Similarly, one can find

cscx,,dx

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§ 8.2 Integration by Parts

Integration by parts is a technique for simplifying integrals of the form

f(x)g(x)dx.

It is useful when f can be differentiated repeatedly and g can be integrated repeatedly without difficulty. For example the integral xexdx and exsin(x)dx can be such type.

Product Rule in Integral Form

Integration by Parts Formula

u(x),v(x),dx=u(x),v(x)u(x),v(x),dx.

The mindset behind the formula is that to transfer a difficult integration to an easier one.

Sometimes it is easier to remember the formula if we write it in differential form.

udv=uvvdu.

Why it works?

The formula for integration by parts is coming from the Product Rule for derivatives

ddx(u(x)v(x))=u(x)v(x)+u(x)v(x).

Example 1. Using Integration by Parts

Find

xcos(x),dx.

Example 3. Integral of the Natural Logarithm

Find

ln(x)dx.

Example 4. Repeated Use of Integration by Parts

Find

x2exdx

Example 5. Solving for the Unknown Integral (An Circling-Back Integration)

Find

excos(x)dx.

Evaluating Definite Integrals by Parts

Combining the Fundamental Theorem of Calculus with the Product Rule, we find the following formula for definite integrals

u(x)v(x)]ab=abu(x)v(x)dx+abu(x)v(x)dx.

Therefore we can find the following integration by parts formula for definite integrals.

Integration by Parts Formula for Definite Integrals

abu(x)v(x)dx=u(x)v(x)]ababu(x)v(x)dx.

Example 6. Finding Area

Find the area of the region bounded by the curve y=xex and the x-axis from x=0 to x=4.

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Example 7. Repeating Multiple Times. Ft. The Tabular Integration Method

Find

x3sin(x)dx.

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Example 9. A Reduction Formula (Iterative Formulas)

Find the integral

In=cosn(x)dx

Example 10. Using the Reduction Formula

Find

cos3(x)dx.