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heureka

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 #3
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integrating: 

dx 2xsin3(x)

 

1. Let

dx sin3(x)=dx sin2(x)sin(x)=dx [1cos2(x)]sin(x)u=cos(x)du=sin(x)dx=du 1sin(x)(1u2)sin(x)=du (1u2)=du+du u2=u+13u3=cos(x)+13cos3(x)

 

2. Let

dx cos3(x)=dx cos2(x)cos(x)=dx [1sin2(x)]cos(x)u=sin(x)du=cos(x)dx=du 1cos(x)(1u2)cos(x)=du (1u2)=dudu u2=u13u3=sin(x)13sin3(x)

 

3. Let

dx xsin3(x)u=xu=1v=cos(x)+13cos3(x)v=sin3(x)=x[cos(x)+13cos3(x)]dx 1[cos(x)+13cos3(x)]=xcos(x)+13xcos3(x)+dx cos(x)dx 13cos3(x)=xcos(x)+13xcos3(x)+sin(x)13dx cos3(x)=xcos(x)+13xcos3(x)+sin(x)13[sin(x)13sin3(x)]=xcos(x)+13xcos3(x)+sin(x)13sin(x)+19sin3(x)]=xcos(x)+13xcos3(x)+23sin(x)+19sin3(x)dx 2xsin3(x)=2xcos(x)+23xcos3(x)+43sin(x)+29sin3(x)=19[18xcos(x)+6xcos3(x)+12sin(x)+2sin3(x)]

 

4. Let

cos3(x)=14[3cos(x)+cos(3x)]sin3(x)=14[3sin(x)sin(3x)]

 

5. Let

dx 2xsin3(x)=19[18xcos(x)+6xcos3(x)+12sin(x)+2sin3(x)]=19{18xcos(x)+6x14[3cos(x)+cos(3x)]+12sin(x)+214[3sin(x)sin(3x)]}=2xcos(x)+16x[3cos(x)+cos(3x)]+43sin(x)+118[3sin(x)sin(3x)]=2xcos(x)+12xcos(x)+16xcos(3x)+43sin(x)+16sin(x)118sin(3x)=32xcos(x)+16xcos(3x)+32sin(x)118sin(3x)

 

or

dx 2xsin3(x)=32xcos(x)+16xcos(3x)+32sin(x)118sin(3x)=118[27xcos(x)+3xcos(3x)+27sin(x)sin(3x)]

 

laugh

05.10.2016