K kr73114 Member Joined Aug 6, 2009 Messages 373 Gender Male HSC 2011 Jan 23, 2011 #1 inegrate: 0-pi/4 [sin2xcos2x.dx] i know you could use reverse chain rule but this question says use the compund angle formula
inegrate: 0-pi/4 [sin2xcos2x.dx] i know you could use reverse chain rule but this question says use the compund angle formula
B Bored Of Fail 2 Banned Joined Jan 13, 2011 Messages 354 Gender Male HSC N/A Jan 23, 2011 #2 kr73114 said: inegrate: 0-pi/4 [sin2xcos2x.dx] i know you could use reverse chain rule but this question says use the compund angle formula Click to expand... use double angle formula for sin2x sin2x= 2sinxcosx therefore sinxcosx = (1/2) sin(2x) therefore sin(2x)cos(2x) = (1/2) sin(4x) now its easy to integrate = -(1/8) [cos(4x)] between the limits etc etc
kr73114 said: inegrate: 0-pi/4 [sin2xcos2x.dx] i know you could use reverse chain rule but this question says use the compund angle formula Click to expand... use double angle formula for sin2x sin2x= 2sinxcosx therefore sinxcosx = (1/2) sin(2x) therefore sin(2x)cos(2x) = (1/2) sin(4x) now its easy to integrate = -(1/8) [cos(4x)] between the limits etc etc
H hscishard Active Member Joined Aug 4, 2009 Messages 2,033 Location study room...maybe Gender Male HSC 2011 Jan 23, 2011 #3 1/2 integral 2sin2xcos2x and then it would be integrate sin4x
