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Summer 2016 MTH142J College Calculus 2
Trigonometric integration
Case
Power of
sin x
Power of
cos x
Solution
(1)
Odd > 0
Any
Substitution u = cos x, du = − sin x d x. Use sin2 x = 1 − cos2 x to transform sine to cosine.
(2)
Any
Odd > 0
(3)
Even ≥ 0
Any
(4)
Any
Even ≥ 0
Substitution u = sin x, du = cos x d x. Use cos2 x = 1 − sin2 x to transform cosine to sine.
R
Use sin2 x = 1 − cos2 x to transform all sine to cosine. Then use reduction formula cosn x d x.
R
Use cos2 x = 1 − sin2 x to transform all cosine to sine. Then use reduction formula sinn x d x.
Z
Z
tan x d x = ln |sec x| + C,
sec x d x = ln |sec x + tan x| + C
Case
Power of
tan x
Power of
sec x
Solution
(5)
Odd > 0
Any
Substitution u = sec x, du = sec x tan x d x. Use tan2 x = sec 2 x − 1 to transform tangent to secant.
(6)
Any
Even > 0
(7)
Even ≥ 0
Any
Substitution u = tan x, du = sec 2 x d x. Use sec 2 x = 1 + tan2 x to transform secant to tangent.
R
Use tan2 x = sec 2 x − 1 to transform all tangent to secant. Then use reduction formula sec n x d x.
Reduction formulas for integrals of trigonometric power:
Z
Z
1
n−1
n−1
sin x d x = − sin
x · cos x +
sinn−2 x d x
n
n
Z
Z
1
n−1
cosn x d x = cosn−1 x · sin x +
cosn−2 x d x
n
n
Z
Z
1
tann−1 x − tann−2 x d x
tann x d x =
n−1
Z
Z
1
n−2
n−2
csc x d x = −
csc
x · cot x +
csc n−2 x d x
n−1
n−1
Z
Z
1
n−2
sec n x d x =
sec n−2 x · tan x +
sec n−2 x d x
n−1
n−1
Z
Z
1
cotn x d x = −
cotn−1 x − cotn−2 x d x
n−1
n
n
Trigonometric values of typical angles:
π
6
= 45◦
π
3
0
sin θ
0
1
2
p
2
2
cos θ
1
p
3
2
p
2
2
1
2
tan θ
0
p
3
3
1
p
3
1
p
2 3
3
p
2
sec θ
= 30◦
π
4
θ
= 60◦
p
3
2
2
π
2
= 90◦
2π
3
= 120◦
3π
4
= 135◦
5π
6
= 150◦
π = 180◦
p
3
2
p
2
2
1
2
0
0
− 12
p
− 22
p
− 23
−1
DNE
p
− 3
−1
−2
p
− 2
1
DNE
−
p
3
3
p
− 233
0
−1
Trigonometric substitution
Form
Substitution
Differential
After transformation
a2 − x 2
x = a sin θ
d x = a cos θ dθ
a2 − x 2 = a2 cos2 θ
x 2 + a2
x = a tan θ
d x = asec 2 θ dθ
x 2 + a2 = a2 sec 2 θ
x 2 − a2
x = asec θ
d x = asec θ tan θ dθ
x 2 − a2 = a2 tan2 θ
Definition of Trigonometric functions:
OP
HP
AJ
cos θ =
HP
OP
sin θ
tan θ =
=
AJ
cos θ
sin θ =
reciprocal
←→
reciprocal
←→
reciprocal
←→
HP
OP
HP
sec θ =
AJ
cos θ
AJ
=
cot θ =
OP
sin θ
csc θ =
Pythagorean theorem: OP2 + AJ2 = HP2 .
Double angle identities
sin2 θ =
1 − cos 2θ
,
2
cos2 θ =
1 + cos 2θ
2
p
a2 − x 2 = a cos θ
p
x 2 + a2 = asec θ
p
x 2 − a2 = a tan θ
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