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[r]

I=

2 2 2

2 2 2 2

0 0 0

3sin 4cos 3 sin 4 cos 3 sin 4 cos

3sin 4cos 3sin 4cos 3sin 4cos cos sin

x x dx x dx x dx x dx x dx

x x x x x x x x

    



   

    

    

Tính A=

2

2

sin cos

x dx x







Đặt t= cosx => dt= -sinxdx

c: Đ

x 2

t

A=

1 03

dt t





Đặt t tan ,u u 2;

         

=>dt 3(tan2u1)du

Đc: t=0=> u=0; t=1 => u=6



A=

2

6

6

0

0

3(tan 3

) |

3(tan 1) 3 18

u u

du du

u

 



 

  



 

Tính B=

2

2

cos sin

x dx x







=

2

2

(sin ) sin

d x

x







=

2

0

(sin )

(sin 2)(sin 2)

d x

x x





 



2

0

1 1

(sin )

4 sinx sinx d x



 

   

    

2

1 sin

ln |

4 sin

x x



     

  

1 ln



Vậy I=

2

2

0

3sin 4cos

3sin 4cos

x x

dx

x x



 



=3A+4B=

3 ln