(1,0 ĐIỂM). Π Π Π+ − − − −2 2 22 2 2X X X X X1 COS 3COS 2SIN 1...

Câu 3 (1,0 điểm).

π

π

π

+ − − − −

2

2

2

x x x x x1 cos 3cos 2sin 1 2sin

∫ ∫ ∫

= ⇔ = − +Ta có ( 2cos )I I x x dx dx+ +x x x x2cos cos

0

0

0

π

π

2

2

2

1 π= − = −  = −

∫

( 2cos ) 2sin 2I x x dx  x x

1

 2 8

0

0

− +

2

2

π

x d x x1 2sin ( 2 cos ) π

∫ ∫

= = = + =ln 2cos lnI dx x x

2

0

2

2cos 2cos 4π

2

πTừ đó ta được 2 lnI = − +8 4