Chapter 6A Notes Page 1 of 12

Introduction and Review

Derivatives 𝑦 = 𝑓(𝑥)

𝑑𝑦

𝑑𝑥= 𝑓′(𝑥)

Evaluate derivative at 𝑥 = 𝑎:

𝑑𝑦

𝑑𝑥|

𝑥=𝑎= 𝑓′(𝑎) = lim

ℎ→0

𝑓(𝑎+ℎ)−𝑓(𝑎)

ℎ

Geometric Interpretation: see figure

slope of the line tangent to 𝑓 at 𝑥 = 𝑎.

Differentiation formulas 𝑑

𝑑𝑥𝑥𝑛 = 𝑛𝑥𝑛−1

𝑑

𝑑𝑥ln|𝑥| =

1

𝑥

𝑑

𝑑𝑥𝑒𝑥 = 𝑒𝑥

𝑑

𝑑𝑥sin(𝑥) = cos(𝑥)

𝑑

𝑑𝑥cos(𝑥) = −sin(𝑥)

𝑑

𝑑𝑥tan(𝑥) = sec2(𝑥)

𝑑

𝑑𝑥sec(𝑥) = sec(𝑥) tan(𝑥)

𝑑

𝑑𝑥sin−1(𝑥) =

1

√1 − 𝑥2

Chapter 6A Notes Page 2 of 12

𝑑

𝑑𝑥tan−1(𝑥) =

1

1 + 𝑥2

Review the sum, difference, product and quotient rules!

The chain rule

Let 𝐹(𝑥) = 𝑓(𝑔(𝑥)), then 𝐹′(𝑥) = 𝑓′(𝑔(𝑥))𝑔′(𝑥)

Or let 𝑦 = 𝑓(𝑢), 𝑢 = 𝑔(𝑥), then 𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢

𝑑𝑢

𝑑𝑥

Example:

𝑑

𝑑𝑥sin(𝑥2) = cos(𝑥2)2𝑥

Indefinite Integrals Indefinite integrals are also known as antiderivatives.

If 𝐹′ is the derivative of 𝐹 then we write

∫ 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶

Example:

∫ cos(𝑥2) 2𝑥 𝑑𝑥 = sin(𝑥2) + 𝐶

Indefinite integral formulas

∫ 𝑥𝑛𝑑𝑥 =1

𝑛 + 1𝑥𝑛+1 + 𝐶

∫1

𝑥 𝑑𝑥 = ln|𝑥| + 𝐶

∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥 + 𝐶

∫ cos(𝑥) 𝑑𝑥 = sin(𝑥) + 𝐶

∫ sin(𝑥) 𝑑𝑥 = − cos(𝑥) + 𝐶

∫ sec2(𝑥)𝑑𝑥 = tan(𝑥) + 𝐶

∫ sec(𝑥) tan(𝑥) 𝑑𝑥 = sec(𝑥) + 𝐶

Chapter 6A Notes Page 3 of 12

Definite Integrals

Example: ∫1

𝑥 𝑑𝑥

𝑒

1

Geometric Interpretation: see figure

Evaluation theorem

∫ 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑥)|𝑎𝑏 = 𝐹(𝑏) − 𝐹(𝑎)

𝑏

𝑎

Examples:

∫1

𝑥 𝑑𝑥 = ln(x)|1

e𝑒

1

= ln(𝑒) − ln(1) = 1 − 0 = 1

∫ cos(𝑥2) 2𝑥 𝑑𝑥 = sin(𝑥2)|01 = sin(1) − sin (0)

1

0

Integration by substitution Suppose an integral has the form: 𝐼 = ∫ 𝑓′(𝑔(𝑥))𝑔′(𝑥)𝑑𝑥,

let 𝑢 = 𝑔(𝑥), then 𝑑𝑢 = 𝑔′(𝑥)𝑑𝑥 (this is differential notation)

Substitute to obtain

𝐼 = ∫ 𝑓′(𝑢)𝑑𝑢 = 𝑓(𝑢) + 𝐶 = 𝑓(𝑔(𝑥)) + 𝐶

This is equivalent to taking the chain rule of differentiation backwards!

Example. 𝐼 = ∫ cos(𝑥2) 𝑥 𝑑𝑥

Let 𝑢 = 𝑥2, 𝑑𝑢 = 2𝑥 𝑑𝑥

𝐼 = ∫ cos(𝑢) 1

2 𝑑𝑢 =

1

2 ∫ cos(𝑢) 𝑑𝑢 =

1

2sin(𝑢) + 𝐶 =

1

2sin(𝑥2) + 𝐶.

Example. 𝐽 = ∫ cos(𝑥2) 𝑥 𝑑𝑥3

2

Chapter 6A Notes Page 4 of 12

Let 𝑢 = 𝑥2. If 𝑥 = 2 then 𝑢 = 4. If 𝑥 = 3 then 𝑢 = 9.

𝐽 = ∫ cos(𝑢)1

2𝑑𝑢

9

4

=1

2sin (𝑢)|4

9

6.1 Integration by parts

Chain rule for differentiation: 𝑑

𝑑𝑥𝑠𝑖𝑛(𝑥2) = (cos(𝑥2))2𝑥

Integration by substitution: let = 𝑥2 , 𝑑𝑢 = 2𝑥 𝑑𝑥

∫ (cos 𝑥2)2𝑥 𝑑𝑥 = ∫ cos(𝑢) 𝑑𝑢 = sin(𝑢) + 𝐶 = sin(𝑥2) + 𝐶

Product rule for differentiation

𝑑

𝑑𝑥[𝑓(𝑥)𝑔(𝑥)] = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)

Solve for𝑓(𝑥)𝑔′(𝑥):

𝑓(𝑥)𝑔′(𝑥) =𝑑

𝑑𝑥[𝑓(𝑥)𝑔(𝑥)] − 𝑓′(𝑥)𝑔(𝑥)

Take antiderivatives

(∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥) − ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥 (A)

This is the rule for integration of indefinite integrals by parts.

Shorthand formula: recall differential notation

Box: 𝑢 = 𝑓(𝑥) 𝑑𝑣 = 𝑔′(𝑥) 𝑑𝑥

𝑑𝑢 = 𝑓′(𝑥)𝑑𝑥 𝑣 = 𝑔(𝑥)

Substitute into (A) above to get

∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 (B)

An easy-to-remember form of the integration by parts formula!

Chapter 6A Notes Page 5 of 12

Example 1. Evaluate 𝐼 = ∫ 𝑥 cos(𝑥) 𝑑𝑥.

Box: 𝑢 = 𝑥 𝑑𝑣 = cos(𝑥) 𝑑𝑥

𝑑𝑢 = 𝑑𝑥 𝑣 = sin (𝑥)

𝐼 = ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢

= 𝑥 sin(𝑥) − ∫ sin (𝑥)𝑑𝑥

= 𝑥 sin(𝑥) + cos(𝑥) + 𝐶

Check:

𝑑

𝑑𝑥(𝑥 sin(𝑥) + cos(𝑥) + 𝐶) = sin(𝑥) + 𝑥 cos(𝑥) − sin(𝑥) = 𝑥 cos (𝑥) √√

In principle, pick 𝑢 so that 𝑑𝑢 is simpler and pick 𝑑𝑣 so that it is possible to integrate. The following

LIPET scheme for choosing 𝑢 and 𝑣 is often useful:

(L)ogarithm

(I)nverse Trig

(P)olynomials or powers of 𝑥

(E)xponential functions

(T)rig functions

Whichever piece of the integrand is higher on the list is 𝑢 and the other piece is 𝑑𝑣.

Example 2. Integration by parts twice

Evaluate 𝐼 = ∫ 𝑥2 sin(𝑥) 𝑑𝑥.

Box: 𝑢 = 𝑥2 𝑑𝑣 = sin(𝑥) 𝑑𝑥

𝑑𝑢 = 2𝑥 𝑑𝑥 𝑣 = −cos (𝑥)

𝐼 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢

= −𝑥2 cos(𝑥) + ∫ cos(𝑥) 2𝑥 𝑑𝑥

= −𝑥2 cos(𝑥) + 2(𝑥 sin(𝑥) + cos(𝑥)) + 𝐶

Chapter 6A Notes Page 6 of 12

We used the result of example 1.

Example 3. Recurrence of the original integral

Evaluate 𝐼 = ∫ 𝑒−𝑥 cos(𝑥) 𝑑𝑥

Box: 𝑢 = 𝑒−𝑥 𝑑𝑣 = cos(𝑥) 𝑑𝑥

𝑑𝑢 = −𝑒−𝑥 𝑑𝑥 𝑣 = sin (𝑥)

𝐼 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 = 𝑒−𝑥 sin(𝑥) + ∫ 𝑒−𝑥 sin(𝑥) 𝑑𝑥 (C)

Evaluate 𝐽 = ∫ 𝑒−𝑥 sin(𝑥) 𝑑𝑥

Box: 𝑢 = 𝑒−𝑥 𝑑𝑣 = sin(𝑥) 𝑑𝑥

𝑑𝑢 = −𝑒−𝑥𝑑𝑥 𝑣 = −cos (𝑥)

𝐽 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 = −𝑒−𝑥 cos(𝑥) − ∫ 𝑒−𝑥 cos(𝑥) 𝑑𝑥 = −𝑒−𝑥 cos(𝑥) − 𝐼 (D)

Combine (C) and (D)

𝐼 = 𝑒−𝑥 sin(𝑥) + (−𝑒−𝑥 cos(𝑥) − 𝐼)

2𝐼 = 𝑒−𝑥(sin(𝑥) − cos(𝑥)) + 2𝐶

𝐼 =1

2𝑒−𝑥(sin(𝑥) − cos(𝑥)) + 𝐶

We added an explicit constant of integration when we put 𝐼 on only one side of the equation. Since 𝐼

represents an indefinite integral, it has an implicit constant of integration.

Example 4: combine parts and substitution

Evaluate 𝐼 = ∫ sin−1(𝑥) 𝑑𝑥

𝑑

𝑑𝑥sin−1(𝑥) =

1

√1−𝑥2

Box: 𝑢 = sin−1(𝑥) 𝑑𝑣 = 𝑑𝑥

𝑑𝑢 =1

√1−𝑥2𝑑𝑥 𝑣 = 𝑥

𝐼 = ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢

= 𝑥 sin−1(𝑥) − ∫𝑥

√1−𝑥2𝑑𝑥

Chapter 6A Notes Page 7 of 12

Evaluate 𝐽 = ∫𝑥

√1−𝑥2𝑑𝑥

Let 𝑡 = 1 − 𝑥2, 𝑑𝑡 = −2𝑥 𝑑𝑥, −1

2𝑑𝑡 = 𝑥 𝑑𝑥

Then 𝐽 = −1

2 ∫ 𝑡−

1

2 𝑑𝑡 = −1

2 (2 𝑡

1

2) + 𝐶 = −𝑡1

2 + 𝐶

Combine to get

𝐼 = 𝑥 sin−1(𝑥) + (1 − 𝑥2)1

2 + 𝐶

Definite Integrals Evaluation Theorem

∫ 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑥)|𝑎𝑏 = 𝐹(𝑏) − 𝐹(𝑎)

𝑏

𝑎

Product Rule

𝑑

𝑑𝑥𝑓(𝑥)𝑔(𝑥) = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)

𝑓(𝑥)𝑔′(𝑥) =𝑑

𝑑𝑥[𝑓(𝑥)𝑔(𝑥)] − 𝑓′(𝑥)𝑔(𝑥)

integrate from 𝑥 = 𝑎 to 𝑥 = 𝑏:

∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥 = ∫𝑑

𝑑𝑥[𝑓(𝑥)𝑔(𝑥)]𝑑𝑥 = ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥

𝑏

𝑎

𝑏

𝑎

𝑏

𝑎

∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥𝑏

𝑎= 𝑓(𝑥)𝑔(𝑥)|𝑎

𝑏 − ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥𝑏

𝑎 (E)

Shorthand using differential notation

Box: 𝑢 = 𝑓(𝑥) 𝑑𝑣 = 𝑔′(𝑥) 𝑑𝑥

𝑑𝑢 = 𝑓′(𝑥)𝑑𝑥 𝑣 = 𝑔(𝑥)

Write as for an indefinite integral:

∫ 𝑢 𝑑𝑣 = 𝑢 𝑣 − ∫ 𝑣 𝑑𝑢

Replace 𝑢, 𝑑𝑢, 𝑣 and 𝑑𝑣 with appropriate expressions in 𝑥 and add limits of integration to get (E)

Chapter 6A Notes Page 8 of 12

Example 1: 𝐼 = ∫ ln(𝑥) 𝑑𝑥𝑒

1

Box: 𝑢 = ln (𝑥) 𝑑𝑣 = 𝑑𝑥

𝑑𝑢 =1

𝑥𝑑𝑥 𝑣 = 𝑥

∫ 𝑢 𝑑𝑣 = 𝑢 𝑣 − ∫ 𝑣 𝑑𝑢

∫ ln(𝑥) 𝑑𝑥 = 𝑥 ln(x)|1e − ∫ dx

e

1

𝑒

1

= (𝑒 ln(𝑒) − 1 ln(1)) − (𝑒 − 1)

= (𝑒 − 0) − (𝑒 − 1)

= 1

Example 2: Combine substitution with parts

𝐼 = ∫ 𝑒√𝑥4

1𝑑𝑥

Let 𝑡 = √𝑥 = 𝑥1

2

𝑑𝑡 =1

2𝑥−

1

2 𝑑𝑥

2 𝑥1

2 𝑑𝑡 = 𝑑𝑥

2𝑡 𝑑𝑡 = 𝑑𝑥

If 𝑥 = 1 then 𝑡 = 1. If 𝑥 = 4 then 𝑡 = 2.

𝐼 = ∫ 𝑒𝑡2𝑡 𝑑𝑡 = 2 ∫ 𝑒𝑡2

1𝑡

2

1 𝑑𝑡

Now use integration by parts

Box: 𝑢 = 𝑡 𝑑𝑣 = 𝑒𝑡𝑑𝑡

𝑑𝑢 = 𝑑𝑡 𝑣 = 𝑒𝑡

∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢

∫ 𝑡𝑒𝑡𝑑𝑡 = 𝑡 𝑒𝑡|12 − ∫ 𝑒𝑡𝑑𝑡

2

1

2

1

= (2𝑒2 − 𝑒) − 𝑒𝑡|12

= (2𝑒2 − 𝑒) − (𝑒2 − 𝑒)

Chapter 6A Notes Page 9 of 12

= 𝑒2

Then 𝐼 = 2𝑒2 is the answer.

6.2A Trigonometric Integrals

Combine trig identities and substitution to evaluate some useful integrals.

Recall the identities:

sin2(𝑥) + cos2(𝑥) = 1 (1)

𝑑

𝑑𝑥sin(𝑥) = cos (𝑥)

𝑑

𝑑𝑥cos(𝑥) = −sin (𝑥)

We will consider two cases of integrals. The first is

∫ sin𝑚(𝑥) cos𝑛(𝑥)𝑑𝑥, (A)

where 𝑚 and 𝑛 are integers.

If 𝑚 or 𝑛 are odd, substitute 𝑢 = sin (𝑥) or 𝑢 = cos (𝑥).

Example. Evaluate 𝐼 = ∫ sin4(𝑥) cos3(𝑥)𝑑𝑥

Cosine is raised to an odd power. Factor out one power of cosine.

𝐼 = ∫ sin4(𝑥) cos2(𝑥) cos(𝑥) 𝑑𝑥

Use (1) to express the rest of the integrand as powers of sine.

𝐼 = ∫ sin4(𝑥)(1 − sin2(𝑥)) cos (𝑥) 𝑑𝑥

Substitute 𝑢 = sin (𝑥), 𝑑𝑢 = cos(𝑥) 𝑑𝑥

𝐼 = ∫ 𝑢4(1 − 𝑢2)𝑑𝑢 = ∫ 𝑢4 − 𝑢6 𝑑𝑢 =1

5𝑢5 −

1

7𝑢7 + 𝐶

𝐼 =1

5sin5(𝑥) −

1

7sin7(𝑥) + 𝐶 ■

Chapter 6A Notes Page 10 of 12

Example. Evaluate 𝐼 = ∫ tan(𝑥) 𝑑𝑥

Recognize 𝐼 = ∫sin (𝑥)

cos (𝑥) dx

Sine is raised to an odd power. Let 𝑢 = cos(𝑥), 𝑑𝑢 = − sin(𝑥) 𝑑𝑥

𝐼 = ∫ −𝑑𝑢

𝑢= − ln|𝑢| + 𝐶 = − ln |cos (𝑥)| + 𝐶 ■

If both 𝑚 and 𝑛 are even in case (A), use the following half angle identities

cos2(𝑥) =1

2(1 + cos(2𝑥)) (2)

sin2(𝑥) =1

2(1 − cos(2𝑥)) (3)

sin(𝑥) cos(𝑥) =1

2sin (2𝑥) (4)

Example. Evaluate 𝐼 = ∫ sin2(𝑥) cos2(𝑥)𝑑𝑥𝜋

20

By (4): sin2(𝑥) cos2(𝑥) =1

4sin2(2𝑥)

By (3): sin2(2𝑥) =1

2(1 − cos(4𝑥))

Thus sin2(𝑥) cos2(𝑥) =1

8(1 − cos(4𝑥))

𝐼 =1

8∫ (1 − cos(4x)) dx

𝜋

20

=1

8∫ 𝑑𝑥

𝜋

20

−1

8∫ cos(4𝑥) 𝑑𝑥

𝜋

20

Let 𝑢 = 4𝑥, 𝑑𝑢 = 4 𝑑𝑥

If 𝑥 = 0 then 𝑢 = 0,

If 𝑥 =𝜋

2 then 𝑢 = 2𝜋.

𝐼 =1

8(

𝜋

2) −

1

8∫ cos(𝑢)

2𝜋

0

1

4𝑑𝑢

where ∫ cos(𝑢)𝑑𝑢 = sin(u)|02π = 0

2𝜋

0

𝐼 =𝜋

16 ■

Chapter 6A Notes Page 11 of 12

Recall the identities

tan2(𝑥) + 1 = sec2(𝑥) (5)

𝑑

𝑑𝑥tan(𝑥) = sec2(𝑥)

𝑑

𝑑𝑥sec(𝑥) = tan(𝑥) sec(𝑥)

The second case of integrals we consider is

∫ tan𝑚(𝑥) sec𝑛(𝑥) 𝑑𝑥 (B)

If 𝑛 is even use the substitution 𝑢 = tan (𝑥)

Example. Evaluate 𝐼 = ∫ tan4(𝑥) sec4(𝑥) 𝑑𝑥

By (5): 𝐼 = ∫ tan4(𝑥)(tan2(𝑥) + 1) sec2(𝑥) 𝑑𝑥

Let 𝑢 = tan (𝑥), 𝑑𝑢 = sec2(𝑥) 𝑑𝑥

𝐼 = ∫ 𝑢4(𝑢2 + 1)𝑑𝑢 = ∫ 𝑢6 + 𝑢4 𝑑𝑢 =1

7𝑢7 +

1

5𝑢5 + 𝐶 =

1

7tan7(𝑥) +

1

5tan5(𝑥) + 𝐶 ■

If 𝑚 is odd use the substitution 𝑢 = sec (𝑥)

Example. Evaluate 𝐼 = ∫ tan3(𝑥) sec3(𝑥) 𝑑𝑥

𝐼 = ∫ tan2(𝑥) sec2(𝑥) tan(𝑥) sec(𝑥) 𝑑𝑥 = ∫ (sec2(𝑥) − 1) sec2(𝑥) tan (𝑥) sec(𝑥) 𝑑𝑥

Let 𝑢 = sec(𝑥), 𝑑𝑢 = tan(𝑥) sec(𝑥) 𝑑𝑥

𝐼 = ∫ (𝑢2 − 1)𝑢2 𝑑𝑢 = ∫ 𝑢4 − 𝑢2 𝑑𝑢 =1

5𝑢5 −

1

3𝑢3 + 𝐶 =

1

5sec5(𝑥) −

1

3sec3(𝑥) + 𝐶 ■

If both 𝑚 is even and 𝑛 is odd, express the integrand entirely in terms of sec (𝑥). This is the hardest case.

Example. Evaluate 𝐼 = ∫ tan2(𝑥) sec(𝑥) 𝑑𝑥

From (5): tan2(𝑥) = sec2(𝑥) − 1

𝐼 = ∫ sec3(𝑥)𝑑𝑥 − ∫ sec(𝑥) 𝑑𝑥

Chapter 6A Notes Page 12 of 12

∫ sec(𝑥) 𝑑𝑥 = ∫ 𝑠𝑒𝑐(𝑥) sec(𝑥)+tan (𝑥)

sec(𝑥)+tan (𝑥) dx

= ∫sec2(𝑥)+sec(𝑥)tan (𝑥)

sec(𝑥)+tan (𝑥) dx

Let 𝑢 = sec(𝑥) + tan(𝑥)

then 𝑑𝑢

𝑑𝑥= tan(𝑥) sec(𝑥) + sec2(𝑥) and 𝑑𝑢 = (tan(𝑥) sec(𝑥) + sec2(𝑥)) 𝑑𝑥

Then ∫ sec(𝑥) 𝑑𝑥 = ∫𝑑𝑢

𝑢= ln|𝑢| + 𝐶 = ln | sec(𝑥) + tan (𝑥)| + 𝐶

(6)

The other integral required to evaluate 𝐼 is treated by example 8 in section 6.2 of our text. The result is:

∫ sec3(𝑥) 𝑑𝑥 =1

2(sec(𝑥) tan(𝑥) + ln | sec(𝑥) + tan(𝑥) | ) + 𝐷 (7)

Results (6) and (7) can be combined to obtain an expression for 𝐼. ■