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MAT01B1: Trig Substitution
Dr Craig
31 July 2018
My details:
I Dr Andrew Craig
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-LAB K02)
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
(C-Les 101, D-Les 105)
Tue 15h30 – 17h05
(D-Les 101, D-Les 203)
I If you have a clash with the tutorial on a
Tuesday, please email me
General information
Saturday class this week:
D1 LAB 108
09h00 to 12h00
Focus on 7.1 – 7.4.
Tough integration by parts question:
Q21:
∫xe2x
(1 + 2x)2dx
Hint: let u = xe2x, dv =1
(1 + 2x)2dx.
Consider the function y =√1− x2.
This represents the top half the unit circle.
1−1
We want to calculate
∫ 1
−1
√1− x2 dx.
But first . . .
POP QUIZ:
Write down the following:
1. the formula for cos(A +B)
2. the three trig squared identities
3. a half-angle formula for cos2 x
4. the formula for sin(A +B)
5. a half-angle formula for sin2 x
6. a formula for sinA cosB
7. the formula for integration by parts
POP QUIZ ANSWERS:
1. cos(A +B) = cosA cosB − sinA sinB
2. sin2 θ + cos2 θ = 1
tan2 θ + 1 = sec2 θ
1 + cot2 θ = csc2 θ
3. cos2 x = 12(1 + cos 2x)
4. sin(A +B) = sinA cosB + sinB cosA
5. sin2 x = 12(1− cos 2x)
6. sinA cosB = 12[sin(A−B)+ sin(A+B)]
7.∫u dv = uv −
∫v du
Remember now the example of∫ 1
−1
√1− x2 dx
Now for a > 0, consider the integral∫ √a2 − x2 dx
(This is a more general case of√1− x2.)
Now let x = a sin θ.
(θ ∈ [−π/2, π/2], more on this later)
Another example:
Consider the integral∫ √a2 + x2 dx
Here we will make a different substitution.
Let x = a tan θ, −π/2 < θ < π/2
For√x2 − a2 we use x = a sec θ.
(Restriction: θ ∈ [0, π/2) ∪ [π, 3π/2).)
Restrictions:
We need to restrict the allowable values of θ
when we make trig substitutions.
Because we are substituting, for example
x = a tan θ
we have to be sure that each value of θ will
produce a unique value for x. Therefore,
we must have that the trig function is
one-to-one on the interval that we allow.
Expression Substitution Restriction
√a2 − x2 x = a sin θ −π
26 θ 6
π
2
√a2 + x2 x = a tan θ −π
2< θ <
π
2
√x2 − a2 x = a sec θ 0 6 θ <
π
2or
π 6 θ <3π
2
Restrictions
The restrictions for trig substitutions must
always be included in your solution.
Also, they are very important when we do
definite integrals.
Example: ∫1
x2√x2 + 4
dx
Solution:
−√x2 + 4
4x+ C
Example:
Evaluate the integral∫ √9− x2x2
dx
Solution:
−√9− x2x
− arcsin(x3
)+ C
Original example: what about bounds?
∫ 1
−1
√1− x2 dx
We let x = sin θ, −π/2 6 θ 6 π/2.
If x = −1, then θ = −π/2.
If x = 1, then θ = π/2.
Note the importance of the restriction on θ.
Without this we would have (infinitely) many
options for θ when −1 = sin θ.
Example:
∫ 3√3/2
0
x3
(4x2 + 9)3/2dx
Solution: note that
(4x2 + 9)3/2 = (√4x2 + 9)3.
You might also need to substitute u = cos θ
at a later stage. Finally you will get
3
32
Example: ellipses
Find the area enclosed by the ellipse
x2
a2+y2
b2= 1
Firstly we see that
1
4A =
∫ a
0
b
a
√a2 − x2 dx
Example:
Evaluate ∫dx√x2 − a2
Solution:
`n|x +√x2 − a2| − `n|a| + C
= `n|x +√x2 − a2| + C1
(Note: hyperbolic trig substitution can also
be used for x > 0.)
Example: ∫x√x2 + 4
dx
Solution: use normal u-substitution to get
√x2 + 4 + C
Example with completing the square:∫x√
3− 2x− x2dx
Solution: complete the square to get
3− 2x− x2 = 4− (x + 1)2.
Substitute u = x + 1. Solution:
−√
3− 2x− x2 − arcsin
(x + 1
2
)+ C
Another example
∫ 2
0
√1 + 4x2 dx
Recall:
∫sec3 x dx =
1
2(secx tanx + ln | secx + tan x|) + C
Solution:∫ 2
0
√1 + 4x2dx =
1
4
(4√17 + ln(
√17 + 4)
)
Preparing for Chapter 7.4:
Practise your long division before Wednesday:
2x3 + 7x2 + 2x + 9
x2 + 3= ?
If you need more revision of long division,
watch the videos using the links provided in
the Blackboard announcement.