Evaluate ∫ex(1+x)cos2xexdx
tan(xex)+c
tan(ex)+c
tan(x+ex)+c
sec(ex)+c
Let z=xex. Then dz=1ex+xexdx=ex1+xdx .
Thus, ∫ex(1+x)cos2xexdx=∫dzcos2z =∫sec2zdz =tanz+c=tanxex+c