groupie skrev:Fra Cambridges opptakstest:
[tex]\int{\sqrt{\tan{(x)}} \rm{d}x[/tex]
Innfør substitusjonen [tex]u = \sqrt{\tan(x)}[/tex]. Da er [tex]x = \arctan(u^2)[/tex], og [tex]\rm{d}x = \frac{2u}{u^4+1} \rm{d}u[/tex]
Da blir integralet
[tex]\int \frac{2u^2}{u^4+1}\rm{d}u = \int \frac{2u^2}{u^4 + 2u^2 + 1 - 2u^2}\rm{d}u = \int \frac{2u^2}{(u^2+1)^2-(\sqrt 2 u)^2}\rm{d}u = \int \frac{2u^2}{(u^2 - \sqrt 2 u + 1)(u^2 + \sqrt 2 u + 1)}\rm{d}u \\ = \frac{1}{2\sqrt 2} \int \left( \frac{u}{u^2 - \sqrt 2 u + 1} - \frac{u}{u^2 + \sqrt 2 u + 1} \right)\rm{d}u[/tex]
Tar de to leddene hver for seg:
[tex]\int \left( \frac{u}{u^2 - \sqrt 2 u + 1} \right) \rm{d}u = \int \left( \frac{u- \sqrt 2}{u^2 - \sqrt 2 u + 1} + \frac{\sqrt 2}{u^2 - \sqrt 2 u + 1} \right) \rm{d}u = \ln |u^2 - \sqrt 2 u + 1| + \int \frac{\sqrt 2}{(u-\frac{1}{\sqrt 2})^2 + (\frac{1}{\sqrt 2})^2} \rm{d}u \\ = \ln |u^2 - \sqrt 2 u + 1| + 2 \arctan(\sqrt 2 u -1) + C[/tex]
[tex]\int \left( \frac{u}{u^2 + \sqrt 2 u + 1} \right) \rm{d}u = \int \left( \frac{u + \sqrt 2}{u^2 + \sqrt 2 u + 1} - \frac{\sqrt 2}{u^2 + \sqrt 2 u + 1} \right) \rm{d}u = \ln |u^2 + \sqrt 2 u + 1| - \int \frac{\sqrt 2}{(u+\frac{1}{\sqrt 2})^2 + (\frac{1}{\sqrt 2})^2} \rm{d}u \\ = \ln |u^2 + \sqrt 2 u + 1| - 2 \arctan(\sqrt 2 u +1) + C[/tex]
Så slår vi dette sammen:
[tex]\int \sqrt{\tan(x)} \rm{d}x = \frac{1}{2\sqrt 2}\ln |\tan(x) - \sqrt{2 \tan(x)} + 1| + \frac{1}{\sqrt 2} \arctan(\sqrt{2 \tan(x)} -1) - \frac{1}{2\sqrt 2}\ln |\tan(x) + \sqrt {2\tan(x)} + 1| +\frac{1}{\sqrt 2} \arctan(\sqrt{2\tan(x)} +1) + C \\ = \frac{1}{2\sqrt 2} \ln \left( \frac{\tan(x) - \sqrt{2 \tan(x)} + 1}{\tan(x) + \sqrt{2 \tan(x)+1} \right) + \frac{1}{\sqrt 2}\arctan(\frac{\sqrt{2 \tan(x)}}{1 - \tan(x) }) +C[/tex]
... Med forbehold om feil.