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sitynightsence
@sitynightsence
July 2022
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NNNLLL54
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Смотри вложение ........................................
3 votes
Thanks 1
dasdasfa
1) =integral(cos^2 (x/2)-2sin(x/2) cos(x/2)+sin^2(x/2)dx=integral(1-sinx)dx=x+cosx+c
2)=integralsin(lnx)d(lnx)=-coslnx+c
3)по частям!
u=x; du=dx
dv=cos^2 xdx; v=integralcos^2 xdx=integral(1+cos(2x))/2 dx=0,5(x+0,5sin(2x))
v=0,5x+0,25sin 2x
integralxcos2 x=x*(0,5x+0,25sin 2x)-integral(0,5x+0,25sin 2x)dx=x*0,5x+0,25sin 2x-
-0,5*(x^2 /2)-0,25/2 *(-cos2x)+c=
4)S=integral(от 0 до 2)! x^3 dx=(x^4) /4 |(от0 до 2)=(2^4)/4-0=16/4=4
0 votes
Thanks 2
sitynightsence
спасибо))
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Answers & Comments
Verified answer
Смотри вложение ........................................2)=integralsin(lnx)d(lnx)=-coslnx+c
3)по частям!
u=x; du=dx
dv=cos^2 xdx; v=integralcos^2 xdx=integral(1+cos(2x))/2 dx=0,5(x+0,5sin(2x))
v=0,5x+0,25sin 2x
integralxcos2 x=x*(0,5x+0,25sin 2x)-integral(0,5x+0,25sin 2x)dx=x*0,5x+0,25sin 2x-
-0,5*(x^2 /2)-0,25/2 *(-cos2x)+c=
4)S=integral(от 0 до 2)! x^3 dx=(x^4) /4 |(от0 до 2)=(2^4)/4-0=16/4=4