24 4 2
dx
x x Warm-up:
24 4 2x x 24 4 1 1x x
22 1 1x
22 1 1
dx
x
Let 2 1u x 2 du dx
2
1
2 1
du
u
11 tan
2u C
11 tan 2 1
2x C
1
2du dx
7.4 Trigonometric Substitutions
Identities you need to know for this section:
2 2
2 2
2 2
Pythagorean Identities:
sin cos 1
tan 1 sec
1 cot csc
2 2
2
2
2
Double Angle Identities:
sin 2 2sin cos
cos 2 cos sin
1 2sin
2cos 1
2 tantan 2
1 tan
Integrals involving Inverse Trig Functions
2 2arcsin
du uC
aa u
2 2
1arctan
du uC
a u a a
2 2
1arcsec
uduC
a au u a
When listing the Antiderivative that corresponds to each of the inverse trigonometric functions, only use one member from each pair.
a is the number.u is the variable.
We can use right triangles and the pythagorean theorem to simplify some problems.
a
x
2 2a x1
24
dx
x
These are in the same form.
2
24 x
24sec
2
x
22sec 4 x
tan2
x
2 tan x
22sec d dx
Write a 'simple' relationship
that involves xKeeping Pyth. Ident.
in mind, write
another 'simple'
relationship
for the other
terms in problem.
We can use right triangles and the pythagorean theorem to simplify some problems.
a
x
2 2a x1
24
dx
x
These are in the same form.
2
24 x
24sec
2
x
22sec 4 x
tan2
x
2 tan x
22sec d dx
22sec
2sec
d
sec d ln sec tan C
24ln
2 2
x xC
We can use right triangles and the pythagorean theorem to simplify some problems.
1
24
dx
x
22sec
2sec
d
sec d ln sec tan C
24ln
2 2
x xC
24ln
2
x xC
2ln 4 ln 2x x C This is a constant.
2ln 4 x x C
a
x
2 2a x
This method is called Trigonometric Substitution.
If the integral contains ,
we use the triangle at right.
2 2a x
If we need , we
move a to the hypotenuse.
2 2a x If we need , we
move x to the hypotenuse.
2 2x a
a
x
2 2a x
a
x2 2x a
2 2
29
x dx
x 3
x
29 x
sin3
x
3sin x
3cos d dx
29cos
3
x
23cos 9 x 29sin 3cos
3cos
d
1 cos 29
2d
91 cos 2
2d
9 9 1sin 2
2 2 2C
sin3
x
1sin3
x
19 9sin 2sin cos
2 3 4
xC
double angle formula
219 9 9
sin2 3 2 3 3
x x xC
2 2
29
x dx
x 3
x
29 x
sin3
x
3sin x
29cos
3
x
23cos 9 x
sin3
x
1sin3
x
19 9sin 2sin cos
2 3 4
xC
double angle formula
219 9 9
sin2 3 2 3 3
x x xC
1 29sin 9
2 3 2
x xx C
3cos d dx
522
dx
x x We can get into the necessary
form by completing the square.
22x x
22x x
2 2 x x
2 2 1 1x x
21 1x
21 1x
21 1
dx
x
Let 1u x
du dx
21
du
u 1
u
21 ucos
cos
d
d C 1 sin u C
1 sin 1x C
21cos
1
u 21 u
sin u cos d du
624 4 2
dx
x x Complete the square:24 4 2x x
24 4 1 1x x
22 1 1x
22 1 1
dx
x
Let 2 1u x 2 du dx
2
1
2 1
du
u
1
u
2 1u
2
2
1 sec
2 sec
d
1
2d
1
2C 11
tan2
u C
11 tan 2 1
2x C
tan u
2sec 1u 1
2du dx
2sec d du
2 2sec 1u
Here are a couple of shortcuts that are result from Trigonometric Substitution:
12 2
1tan
du uC
u a a a
1
2 2sin
du uC
aa u
These are on your list of formulas. They are not really new.
p
HW Day 1: p. 512 #’s 1-4, 5-17 odd, 41-45 odd
In Class/HW Day 2: p. 512 #’s 19-37 odd, 47, 49 odd
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