S _ShiFTy_ Member Joined Aug 7, 2005 Messages 185 Gender Male HSC 2006 Apr 4, 2006 #1 ∫sinx.cosx.e^cos2x dx I have a mental blank for this question...the cos2x is really bugging me I tried letting u = sinx.cosx, giving du = cos^2(x) - sin^2(x) = cos2x But that made absolutely no sense Last edited: Apr 4, 2006
∫sinx.cosx.e^cos2x dx I have a mental blank for this question...the cos2x is really bugging me I tried letting u = sinx.cosx, giving du = cos^2(x) - sin^2(x) = cos2x But that made absolutely no sense
S _ShiFTy_ Member Joined Aug 7, 2005 Messages 185 Gender Male HSC 2006 Apr 4, 2006 #2 Ahh..silly me, thanks for that
V vafa Member Joined Mar 16, 2006 Messages 302 Gender Undisclosed HSC N/A Apr 4, 2006 #3 let u=cos2x du/dx=-2sin2x=-4sinxcosx S indicates integration sign S sinxcosx.e^(cos2x)=-1/4 S -4sinx.cosx.e^cos2x=-1/4 S du/dx.e^u.dx when u multiply du/dx.dx dxs cancel out each other so u will get: -1/4 S e^u du=-1/4 e^u+c=-1/4e^cos2x+c
let u=cos2x du/dx=-2sin2x=-4sinxcosx S indicates integration sign S sinxcosx.e^(cos2x)=-1/4 S -4sinx.cosx.e^cos2x=-1/4 S du/dx.e^u.dx when u multiply du/dx.dx dxs cancel out each other so u will get: -1/4 S e^u du=-1/4 e^u+c=-1/4e^cos2x+c
S _ShiFTy_ Member Joined Aug 7, 2005 Messages 185 Gender Male HSC 2006 Apr 4, 2006 #4 Oh no wonder...i kept on treating cos2x as a power, eesh..silly me
