Home
О нас
Products
Services
Регистрация
Войти
Поиск
mkineshma
@mkineshma
July 2022
1
2
Report
интеграл
[tex] \int\limits \, \frac{dx}{asinx+3cosx} [/tex]
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms of service
You must agree before submitting.
Send
Answers & Comments
Gerren
Используем подстановку
u=tg(x/2) du=(1/2)dx/(sin(x/2))^2, тогда
sinx=2u/(u^2+1) cosx=(1-u^2)/(u^2+1) dx=2du/(u^2+1), получили
∫2du/((u^2+1)(2au/(u^2+1)+3(1-u^2)/(u^2+1)), упростив знаменатель, имеем
∫2du/(-3u^2+2au+3)=2∫du/(-3u^2+2au+3)=2∫(1/12)(36+4a^2)-(√3u-a/√3)^2)
s=√3u-pi/3 ds=√3du, получим
2/√3∫12ds/(36+4a^2)-s^2)=8√3/(36+4a^2)∫ds/(1-3s^2/(9+a^2))
Снова заменим p=√(3/(a^2+9) dp=√((3/(a^2+9))ds
8√(9+a^2)/(36+4a^2)∫dp/(a-p^2)=8√(9+a^2)Arth(p)/(36+4a^2)+С
Делаем возврат
8√(9+a^2)Arth(p)/(36+4a^2)+С=2Arth(√(3s/(a^2+9))/(9+a^2)+C=2Arth(3u-a)/(√(9+a^2))/√(9+a^2)+C=2Arth(3tg(x/2)-a)/(√(9a^2))/√(9+a^2)+C
1 votes
Thanks 1
More Questions From This User
See All
mkineshma
August 2022 | 0 Ответы
tex]...
Answer
mkineshma
August 2022 | 0 Ответы
((e^2x)+3)...
Answer
mkineshma
August 2022 | 0 Ответы
y log 3x 1 postroit tablicu xy
Answer
mkineshma
August 2022 | 0 Ответы
tex]...
Answer
mkineshma
August 2022 | 0 Ответы
najti tochki razryva funkcii i provesti ih klassifikaciyu yx4lnx42b4a75a8f20114f9f3b3dcd26aabb2235 18878
Answer
mkineshma
August 2022 | 0 Ответы
tex]...
Answer
mkineshma
August 2022 | 0 Ответы
tex]...
Answer
mkineshma
July 2022 | 0 Ответы
texferencialnoj proverkoj
Answer
mkineshma
July 2022 | 0 Ответы
pomogite postorit prosto chertezh po zadache pryamougolnyj treugolniks katetami
Answer
mkineshma
July 2022 | 0 Ответы
tex]...
Answer
рекомендуемые вопросы
rarrrrrrrr
August 2022 | 0 Ответы
o chem dolzhny pozabotitsya v pervuyu ochered vzroslye pri organizacionnom vyvoze n
danilarsentev
August 2022 | 0 Ответы
est dva stanka na kotoryh vypuskayut odinakovye zapchasti odin proizvodit a zapcha
myachina8
August 2022 | 0 Ответы
najti po grafiku otnoshenie v3v1 v otvetah napisano 9 no nuzhno reshenie
ydpmn7cn6w
August 2022 | 0 Ответы
Choose the correct preposition: 1.I am fond (out,of,from) literature. 2.where ar...
millermilena658
August 2022 | 0 Ответы
opredelite kak sozdavalas i kto sozdaval arabskoe gosudarstvo v kracii
MrZooM222
August 2022 | 0 Ответы
ch ajtmanov v rasskaze krasnoe yabloko ispolzuet metod rasskaz v rasskaze opi
timobila47
August 2022 | 0 Ответы
kakovo bylo naznachenie kazhdoj iz chastej vizantijskogo hrama pomogite pozhalujsta
ivanyyaremkiv
August 2022 | 0 Ответы
moment. 6....
pozhalujsta8b98a56c0152a07b8f4cbcd89aa2f01e 97513
sarvinozwakirjanova
August 2022 | 0 Ответы
pomogite pozhalusto pzha519d7eb8246a08ab0df06cc59e9dedb 6631
×
Report "tex]..."
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
О нас
Политика конфиденциальности
Правила и условия
Copyright
Контакты
Helpful Social
Get monthly updates
Submit
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
u=tg(x/2) du=(1/2)dx/(sin(x/2))^2, тогда
sinx=2u/(u^2+1) cosx=(1-u^2)/(u^2+1) dx=2du/(u^2+1), получили
∫2du/((u^2+1)(2au/(u^2+1)+3(1-u^2)/(u^2+1)), упростив знаменатель, имеем
∫2du/(-3u^2+2au+3)=2∫du/(-3u^2+2au+3)=2∫(1/12)(36+4a^2)-(√3u-a/√3)^2)
s=√3u-pi/3 ds=√3du, получим
2/√3∫12ds/(36+4a^2)-s^2)=8√3/(36+4a^2)∫ds/(1-3s^2/(9+a^2))
Снова заменим p=√(3/(a^2+9) dp=√((3/(a^2+9))ds
8√(9+a^2)/(36+4a^2)∫dp/(a-p^2)=8√(9+a^2)Arth(p)/(36+4a^2)+С
Делаем возврат
8√(9+a^2)Arth(p)/(36+4a^2)+С=2Arth(√(3s/(a^2+9))/(9+a^2)+C=2Arth(3u-a)/(√(9+a^2))/√(9+a^2)+C=2Arth(3tg(x/2)-a)/(√(9a^2))/√(9+a^2)+C