[tex]\int{\sin \sqrt{x}dx}[/tex]
Har jobbe med denne oppgaven en del, har kommet frem til ;
[tex]\frac{\sin^2 \sqrt{x}}{2}+C[/tex]
Mens i fasitten står det;
[tex]2(\sin \sqrt{x} - \sqrt{x}\cos \sqrt{x})+C[/tex]
Enda ett ubestemt integral
Moderators: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]u = \sqrt{x} \ \Rightarrow \ \frac{\rm{d}u}{\rm{d}x} = \frac{1}{2\sqrt{x}} \ \Rightarrow \ \rm{d}x = 2u\rm{d}u[/tex]
[tex]\int\sin{\sqrt{x}}\rm{d}x = 2\int u\sin{u}\rm{d}u[/tex]
[tex]2\int u\sin{u}\rm{d}u = 2(-u\cos{u} + \int\cos{u}\rm{d}u) = 2(-u\cos{u} + \sin{u}) + C = 2(\sin{\sqrt{x} - \sqrt{x}\cos{\sqrt{x}}) + C[/tex]
[tex]\int\sin{\sqrt{x}}\rm{d}x = 2\int u\sin{u}\rm{d}u[/tex]
[tex]2\int u\sin{u}\rm{d}u = 2(-u\cos{u} + \int\cos{u}\rm{d}u) = 2(-u\cos{u} + \sin{u}) + C = 2(\sin{\sqrt{x} - \sqrt{x}\cos{\sqrt{x}}) + C[/tex]