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 1x

xx

d

(sec

dx

1

p1

(x)) = 2

xx

?0 1x

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

n2

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

n1

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

n1

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

n2

n 2R n 2

sec

(x) =

n

1 tan(x) sec

(x) +

n

1

sec

(x)dx

R

dx

1

1x

2+ 2 = tan

+C

xa a

a

R

1

n

n2

n 2R n 2

R

p dx

1

1x

csc (x) = 1 cot(x) csc (x) + 1 csc (x)dx

2 2 = sec

+C

n

n

xx a a

a

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Tel: @csusm_stemcenter STEM SC (N): (760) 750-4101

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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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