MAT01B1: Trigonometric Integrals

Dr Craig

26 July 2017

My details:

I Dr Andrew Craig

I acraig@uj.ac.za

I Consulting hours:

Monday 14h40 – 15h25

Thursday 11h20 – 12h55

Friday 11h20 – 12h55

I Office C-Ring 508

https://andrewcraigmaths.wordpress.com/

or google “Andrew Craig maths”

General information

I Lectures:

Mon 15h30 – 17h05 (D-Les 201)

Tue 08h50 – 10h25 (D-Les 101)

Mon & Tue lectures cover the same

topics.

Wed 17h10 – 18h45

(D-Les 101 and D-Les 102)

General information

I Tutorials:

Tue 13h50 – 15h25

(D-Les 105, D-Les 203)

Tue 15h30 – 17h05

(D-Les 203, D-Les 204)

I If you have a clash with the tutorial on a

Tuesday, please email me

(acraig@uj.ac.za).

General information

Saturday class this week:

D1 LAB 108 & 109

09h00 to 12h00

Focussing on u-substitution, integration by

parts, trigonometric integrals.

General information

I MAT01B1 will be harder than

MAT01A1.

I You will need to put in more hours perweek in order to succeed in this course.

I You need at least 8 hours per weekoutside of class time.

Today we will learn how to handle integrals

of the form:

I

∫sinm x cosn x dx

I

∫tanm x secn x dx

I

∫cotm x cscn x dx

And of the form:

I

∫sinmx cosnx dx

I

∫sinmx sinnx dx

I

∫cosmx cosnx dx

POP QUIZ:

Write down the following:

1. the formula for integration by parts

2. the three trig squared identities

3. the formula for sin(A +B)

4. the formula for cos(A +B)

POP QUIZ ANSWERS:

1.∫u dv = uv −

∫v du

2. sin2 θ + cos2 θ = 1

tan2 θ + 1 = sec2 θ

1 + cot2 θ = csc2 θ

3. sin(A +B) = sinA cosB + sinB cosA

4. cos(A +B) = cosA cosB − sinA sinB

First examples of trig integrals:

∫cos3 x dx = sinx− 1

3sin3 x + C

∫sin5 x cos2 x dx

=−13

cos3 x +2

5cos5 x− 1

7cos7 x + C

Even powers of sinx and cosx

For these we use the half-angle identities

sin2 x =1

2(1− cos 2x)

cos2 x =1

2(1 + cos 2x)

Example: evaluate∫ π

0

sin2 x dx =π

2

Even powers of sinx and cosx

sin2 x =1

2(1− cos 2x)

cos2 x =1

2(1 + cos 2x)

Another example: evaluate∫sin4 x dx

=1

4

(3

2x− sin 2x +

1

8sin 4x

)+ C

Strategy for∫sinm x cosn x dx

1. If the power of cosx is odd, save one

cosx and convert the remaining cosx

factors to terms involving sinx.

2. If the power of sinx is odd, save one

sinx and convert the remaining sinx

factors to terms involving cosx.

3. If both powers are even, use the

half-angle identities.

Integrals with tanx and secx∫tan6 x sec4 x dx∫tan5 θ sec7 θ dθ

Strategy for∫tanm x secn x dx

1. If the power of secx is even, save sec2 x

and convert the remaining secx to terms

involving tanx.

2. If the power of tanx is odd, save

secx tanx and convert the remaining

tanx to terms involving secx.

What about∫sec4 x tan3 x dx ?

Blue =sec6 x

6− sec4 x

4Red =tan4 x

4+

tan6 x

6=

sec6 x

6− sec4 x

4+

1

12

When we have∫sinm x cosn x dx

with both m and n odd we can choose which

method to use. For∫tanm x secn x dx

with m odd and n even we can also choose

which approach to take.

Exercise to do at home:

Experiment and come up with your own rules

for integrals of the form:∫cotn x cscm x dx

Test your rules on exercises 35 – 40.

For other cases, the guidelines are not

as clear-cut. We may need to use

identities, integration by parts, and

occasionally a little ingenuity.

Harder integrals with secx and tanx

Observe the following:∫secx dx =

∫ (secx× secx + tanx

secx + tanx

)dx

= ln | secx + tan x| + C

and recall that∫tanx dx = ln | secx| + C

Harder integrals with secx and tanx

Use the integrals on the previous slide to

show that:∫tan3 x dx =

tan2 x

2− ln | secx| + C

∫sec3 x dx =

1

2(secx tanx + ln | secx + tan x|) + C

Depending which choice you made in the

integration, you might get∫tan3 x dx =

sec2 x

2− ln | secx| + C.

For∫sec3 x dx, you will need to use

integration by parts.

Simplify the following expressions using the

addition and subtraction formulas:

sin(A−B) + sin(A +B)

cos(A−B)− cos(A +B)

cos(A−B) + cos(A +B)

Hence derive formulas for the following:

sinA cosB sinA sinB cosA cosB

We get:

1. sinA cosB = 12[sin(A−B)+ sin(A+B)]

2. sinA sinB = 12[cos(A−B)− cos(A+B)]

3. cosA cosB = 12[cos(A−B)+cos(A+B)]

We can use the above formulas to solve

integrals of the form:

1.∫sinmx cosnx dx

2.∫sinmx sinnx dx

3.∫cosmx cosnx dx

Example: ∫sin 4x cos 5x dx

Solution:

1

2

(cosx− 1

9cos 9x

)+ C