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mkineshma
@mkineshma
July 2022
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[tex] \int\limits \, \frac{dx}{asinx+3cosx} [/tex]
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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
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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