B bos1234 Member Joined Oct 9, 2006 Messages 491 Gender Male HSC 2007 Jun 22, 2007 #1 how do i integrate xcos2x sec^4 x?
M Mattamz Member Joined Aug 13, 2005 Messages 64 Gender Male HSC 2007 Jun 22, 2007 #2 i dont think they would give you a question like this in 2u, without something like... Find: a) d/dx [(xsin2x)/2+(cos2x)/4} b) Hence find Int(xcos2x)dx if u do 4u maths could could do intergration by parts ie. let v = x u = (sin2x)/2 dv=dx du = cos2x.dx Int(xcos2x)dx = (xsin2x)/2 - Int((sin2x)/2)dx =(xsin2x)/2 + (cos2x)/4 + C
i dont think they would give you a question like this in 2u, without something like... Find: a) d/dx [(xsin2x)/2+(cos2x)/4} b) Hence find Int(xcos2x)dx if u do 4u maths could could do intergration by parts ie. let v = x u = (sin2x)/2 dv=dx du = cos2x.dx Int(xcos2x)dx = (xsin2x)/2 - Int((sin2x)/2)dx =(xsin2x)/2 + (cos2x)/4 + C
S SoulSearcher Active Member Joined Oct 13, 2005 Messages 6,757 Location Entangled in the fabric of space-time ... Gender Male HSC 2007 Jun 22, 2007 #3 For the second question, convert sec4x to sec2x * sec2x, then convert one of the sec2x terms into 1 + tan2x, expand the brackets and then integrate. You might need the substitution u = tan x for one of those terms.
For the second question, convert sec4x to sec2x * sec2x, then convert one of the sec2x terms into 1 + tan2x, expand the brackets and then integrate. You might need the substitution u = tan x for one of those terms.