Integration Problem (1 Viewer)

[_ShiFTy_]

∫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

 

[_ShiFTy_]

Ahh..silly me, thanks for that

 

[vafa]

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

 

[_ShiFTy_]

Oh no wonder...i kept on treating cos2x as a power, eesh..silly me