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iMichael
@iMichael
August 2022
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Интегрировать тригонометрическое выражение.
Осталось решить только эти примеры, кто может, пожалуйста помогите.
№1. ∫4[tex] sin^{2} [/tex]3xdx.
№2.∫[tex] sin^{3}[/tex]3xdx.
№3.∫(1+[tex]cos^{2}[/tex]x)dx.
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sedinalana
Verified answer
1
4sin²3x=4*(1-cos6x)/2=2(1-cos6x)=2-2cos6x
S(2-2cos6x)dx=S2dx-2Scos6xdx=2x-1/3*sin6x+C
2
1+cos²x=1+(1+cos2x)/2=1+1/2+1/2*cos2x=1,5+0,5cos2x
S(1,5+0,5cos2x)dx=S1,5dx+0,5Scos2xdx=1,5x+0,25sin2x+C3
3
Ssin³3xdx=-1/3*cos3xsin²3x+2/3Ssin²3xdx=
=-1/3*cos3xsin²3x+2/3S(1-cos6x)2dx=
=-1/3*cos3xsin²3x+1/3Sdx-1/3Scos6xdx=
=-1/3*cos3xsin²3x+1/3*x-1/18*sin6x+C
3 votes
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Answers & Comments
Verified answer
14sin²3x=4*(1-cos6x)/2=2(1-cos6x)=2-2cos6x
S(2-2cos6x)dx=S2dx-2Scos6xdx=2x-1/3*sin6x+C
2
1+cos²x=1+(1+cos2x)/2=1+1/2+1/2*cos2x=1,5+0,5cos2x
S(1,5+0,5cos2x)dx=S1,5dx+0,5Scos2xdx=1,5x+0,25sin2x+C3
3
Ssin³3xdx=-1/3*cos3xsin²3x+2/3Ssin²3xdx=
=-1/3*cos3xsin²3x+2/3S(1-cos6x)2dx=
=-1/3*cos3xsin²3x+1/3Sdx-1/3Scos6xdx=
=-1/3*cos3xsin²3x+1/3*x-1/18*sin6x+C