[tex]u = \ln x \\ \, \\ x = e^u \\ \, \\ [/tex]Janhaa wrote:Mr MathNoob, prøv deg på denne:
[tex]I=\int \sin(\ln(x))\,{\rm dx}[/tex]
[tex]dx = e^u \, du[/tex]
[tex]I = \int e^u \sin(u) \rm{du} = e^u\sin(u) - \int e^u \cos(u)\rm{du} \\ \, \\ I = \int e^u \sin(u) \rm{du} = e^u\sin(u) - e^u \cos(u) - \int e^u \sin(u)\rm{du} \\ \, \\ I = \int \sin(\ln x) \rm{dx} = \frac 12x \left(\sin(\ln x) - \cos(\ln x)\right) + \rm{C} [/tex]
Blir det bra?