Indefinite integral of sin 2x dx

Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter58: Achievement Review—section Five
Section: Chapter Questions
Problem 30AR: Determine dimension x to 3 decimal places.
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Please answer number 5, evaluate indefinite integral, put the complete solution, and box the final answer.
5. Indefinite integral of sin 2x dx
Transcribed Image Text:5. Indefinite integral of sin 2x dx
Garcia College
Kalibo, Aklan
LECTURE NOTES IN MATH 121 (INTEGRAL CALCULUS)
INSTRUCTOR: Engr. Josefina R. Dagohoy, MEN9
NAME OF STUDENT
PROGRAM & YEAR
INDEFINITE INTEGRALS (Trigonometric Functions)
Basic Trigonometric Derivatives and their Corresponding Indefinite Integrals
sin u = cos u
du
S cos u du = sin u + C
cos u = -sin u
f sin u du =- cos u + C
du
tan u = sec²u
du
( sec u du = tan u + C
cot u = -csc²u
du
S csc?u du =- cot u + C
sec u = sec u tan uL
du
S sec u tan u du = sec u+C
csc u = -csc u cot u
du
ſ csc u cot u du = -csc u+C
Notice the way in which the functions pair off for purposes of integration. The
sin u and cos u, sec u and tan u, csc u and cot u fit well together. An integral
ring, for instance, sin u and tanu is not in appropriate form for application of simple
ration formulas. Upon meeting such an integral we first put the integrand entirely
of sin u and cos u or in terms of tan u sec u.
miprocal Identities:
Pythagorean Identities:
sin? 0+cos?0 = 1
oso e-
sin e
Cso e
1
sco 8-
1+ tan? e = sec? 0
%3D
scc 8
cos e
1+cot?0 = csc?0
%3D
cot 0-
tan 6
cot e
Even Odd Identities:
sin(-0)=-sin 6, cse(-0)= -cac @
Transcribed Image Text:Garcia College Kalibo, Aklan LECTURE NOTES IN MATH 121 (INTEGRAL CALCULUS) INSTRUCTOR: Engr. Josefina R. Dagohoy, MEN9 NAME OF STUDENT PROGRAM & YEAR INDEFINITE INTEGRALS (Trigonometric Functions) Basic Trigonometric Derivatives and their Corresponding Indefinite Integrals sin u = cos u du S cos u du = sin u + C cos u = -sin u f sin u du =- cos u + C du tan u = sec²u du ( sec u du = tan u + C cot u = -csc²u du S csc?u du =- cot u + C sec u = sec u tan uL du S sec u tan u du = sec u+C csc u = -csc u cot u du ſ csc u cot u du = -csc u+C Notice the way in which the functions pair off for purposes of integration. The sin u and cos u, sec u and tan u, csc u and cot u fit well together. An integral ring, for instance, sin u and tanu is not in appropriate form for application of simple ration formulas. Upon meeting such an integral we first put the integrand entirely of sin u and cos u or in terms of tan u sec u. miprocal Identities: Pythagorean Identities: sin? 0+cos?0 = 1 oso e- sin e Cso e 1 sco 8- 1+ tan? e = sec? 0 %3D scc 8 cos e 1+cot?0 = csc?0 %3D cot 0- tan 6 cot e Even Odd Identities: sin(-0)=-sin 6, cse(-0)= -cac @
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ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,