integral sinx sin2x sin3x dx

1.	integral sinx sin2x sin3x dx
Answer» ..= (1/2) ∫ [ 2 sin 2x. sin x ] sin 3x dx ..= (1/2) ∫ [ cos ( 2x-x) - cos ( 2x+x) ] sin 3x dx ..= (1/2) ∫ ( cos x - cos 3x ) sin 3x dx ..= (1/4) ∫ [ ( 2 sin 3x cos x ) - ( 2 sin 3x cos 3x ) ] dx ..= (1/4) ∫ { [ sin ( 3x + x ) + sin ( 3x - x ) ] - ( sin 2·3x ) } dx ..= (1/4) ∫ [ ( sin 4x + sin 2x ) - ( sin 6x ) ] dx ..= (/4) { [ ( - cos 4x ) / 4 ] + [ ( - cos 2x ) / 2 ] - [ ( - cos 6x ) / 6 ] } + C ..= ( -1/16) cos 4x - (1/8) cos 2x + (1/24) cos 6x + C ................................ Ans.

Answer»

..= (1/2) ∫ [ 2 sin 2x. sin x ] sin 3x dx

..= (1/2) ∫ [ cos ( 2x-x) - cos ( 2x+x) ] sin 3x dx

..= (1/2) ∫ ( cos x - cos 3x ) sin 3x dx

..= (1/4) ∫ [ ( 2 sin 3x cos x ) - ( 2 sin 3x cos 3x ) ] dx

..= (1/4) ∫ { [ sin ( 3x + x ) + sin ( 3x - x ) ] - ( sin 2·3x ) } dx

..= (1/4) ∫ [ ( sin 4x + sin 2x ) - ( sin 6x ) ] dx

..= (/4) { [ ( - cos 4x ) / 4 ] + [ ( - cos 2x ) / 2 ] - [ ( - cos 6x ) / 6 ] } + C

..= ( -1/16) cos 4x - (1/8) cos 2x + (1/24) cos 6x + C ................................ Ans.

Discussion