CALCULUS TRIGONOMETRIC DERIVATIVES AND INTEGRALS
CALCULUS
TRIGONOMETRIC DERIVATIVES AND INTEGRALS
TRIGONOMETRIC DERIVATIVES
d
? 0
(sin(x)) = cos(x) x
dx
d
? 0
(cos(x)) = sin(x) x
dx
d
? 0
d
? 0
(csc(x)) = csc(x) cot(x) x
(sec(x)) = sec(x) tan(x) x
dx
dx
d
(sin
dx
1
p1
(x)) = 1
2
? 0 x
x
d
1
(cos (x)) =
dx
p1 1
2
? 0 x
x
d
1
(csc (x)) =
dx
p1
2
? 0 1 x
xx
d
(sec
dx
1
p1
(x)) = 2
xx
? 0 1 x
d
2 ?0
(tan(x)) = sec (x) x
dx
d
2 ?0
(cot(x)) = csc (x) x
dx
d
(tan
1
(x))
=
1 1+
2
? 0 x
dx
x
d
1
(cot (x)) =
1 1+
2
? 0 x
dx
x
TRIGONOMETRIC INTEGRALS R sin(x)dx = cos(x) + C
R cos(x)dx = sin(x) + C
R tan(x)dx = ln | sec(x)| + C
R
|
|
csc(x)dx = ln csc(x) cot(x) + C
R sec(x)dx = ln | sec(x) + tan(x)| + C
R cot(x)dx = ln | sin(x)| + C
POWER REDUCTION FORMULAS
R
n
1
1 n
n 1R
2 n
sin (x) = sin (x) cos(x) +
sin (x)dx
n
n
INVERSE TRIG INTEGRALS
R
1
1
p
2
sin (x)dx = x sin (x) + 1 x + C
R
n
1 n1
n 1R
n 2
cos (x) = cos (x) sin(x) +
cos (x)dx
n
n
R
1
p
1
2
cos (x)dx = x cos (x) 1 x + C
R
n
1
n 1
R n2
tan (x) = 1 tan (x) tan (x)dx
n
R
1
1
1
2
tan (x)dx = x tan (x) 2 ln(1 + x ) + C
R
n
1
n 1
R n2
cot (x) = 1 cot (x) cot (x)dx
n
R
p dx
1 x
2 2 = sin
+ C
ax
a
R
1
n
n 2
n 2R n 2
sec
(x) =
n
1 tan(x) sec
(x) +
n
1
sec
(x)dx
R
dx
1
1 x
2+ 2 = tan
+ C
xa a
a
R
1
n
n 2
n 2R n 2
R
p dx
1
1 x
csc (x) = 1 cot(x) csc (x) + 1 csc (x)dx
2 2 = sec
+ C
n
n
xx a a
a
csusm.edu/stemsc
XXX
Tel: @csusm_stemcenter STEM SC (N): (760) 750-4101
STEM SC (S): (760) 750-7324
CALCULUS
TRIGONOMETRIC DERIVATIVES AND INTEGRALS
R
STRATEGY FOR EVALUATING m
n
sin (x) cos (x)dx
2
2
(a) If the power n of cosine is odd (n = 2k + 1), save one cosine factor and use cos (x) = 1 sin (x) to
express the rest of the factors in terms of sine:
Z
Z
Z
m
n
m
2k+1
m
2k
sin (x) cos (x)dx = sin (x) cos (x)dx = sin (x)(cos (x)) cos(x)dx
Z
2
m
k
= sin (x)(1 sin (x)) cos(x)dx
Then solve by u-substitution and let u = sin(x).
2
2
(b) If the power m of sine is odd (m = 2k + 1), save one sine factor and use sin (x) = 1 cos (x) to
express the rest of the factors in terms of cosine:
Z
Z
Z
m
n
2k+1
n
2 kn
sin (x) cos (x)dx = sin (x) cos (x)dx = (sin (x)) cos (x) sin(x)dx
Z
2 kn
= (1 cos (x)) cos (x) sin(x)dx
Then solve by u-substitution and let u = cos(x).
(b) If both powers m and n are even, use the half-angle identities:
2
1
sin (x) = (1 cos(2x))
2
2
1
cos (x) = (1 + cos(2x))
2
R
STRATEGY FOR EVALUATING
m
n
tan (x) sec (x)dx
2
2
2
(a) If the power n of secant is even (n = 2k, k 2), save one sec (x) factor and use sec (x) = 1 + tan (x)
to express the rest of the factors in terms of tangent:
Z
Z
Z
m
n
2
m
k
212
m
k
tan (x) sec (x)dx = tan (x) sec (x)dx = tan (x)(sec ) sec (x)(x)dx
Z
2
12
m
k
= tan (x)(1 + tan (x)) sec (x)(x)dx
Then solve by u-substitution and let u = tan(x).
2
(b) If the power m of tangent is odd (m = 2k + 1), save one sec(x) tan(x) factor and use tan (x) =
2
sec (x) 1 to express the rest of the factors in terms of secant:
Z
Z
Z
m
n
2k+1
n
2 k n1
tan (x) sec (x)dx = tan (x) sec (x)dx = (tan (x)) sec (x) sec(x) tan(x)dx
Z
2
k n1
= (sec (x) 1) sec (x) sec(x) tan(x)dx
Then solve by u-substitution and let u = sec(x).
csusm.edu/stemsc
XXX
Tel: @csusm_stemcenter STEM SC (N): (760) 750-4101
STEM SC (S): (760) 750-7324
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