TrulsBR wrote:Vi kan ikke gi oss med et slikt integral, så her er, tradisjonen tro, et nytt nattintegral:
[tex]I=\int {\frac{1}{{\cos ^8 x - \sin ^8 x}}\rm{d}x}[/tex]
Jeg prøver - med forbehold om feil...
[tex]I=\int \frac{\sec^8(x)}{1-\tan^8(x)}{\rm dx}=\int \frac{(1+\tan^2(x))^4}{1-\tan^8(x)}{\rm dx}=\int \frac{(1+\tan^2(x))^2(1+\tan^2(x))^2}{1-\tan^8(x)}{\rm dx}[/tex]
u = tan[sup]2[/sup](x) ==> du = 2tan(x)*(1 + tan[sup]2[/sup](x)) dx
altså:
[tex]I={1\over 2}\int \frac{(1+u)^3}{\sqrt{u} (1-u^4)} {\rm du}[/tex]
s = [symbol:rot]u
[tex]I=\int \frac{(s^2+1)^3}{1-s^8} {\rm ds}=-\int \frac{(s^2+1)^2}{(s+1)(s-1)(s^2+s sqrt{2}+1)(s^2-s\sqrt{2}+1)}{\rm ds}[/tex]
delbrøksoppspalter:
[tex]I=- \int (\frac{s\sqrt{2}+1}{2(s^2+s\sqrt{2}+1)}\,+\,\frac{s\sqrt{2}-1}{2(-s^2s+\sqrt{2}-1)}\,-\,\frac{1}{s+1}\,+\,\frac{1}{s-1})\,{\rm ds}[/tex]
[tex]I=\ln|\frac{s-1}{s+1}|\,+\,{1\over 2}(\int \frac{s \sqrt{2}-1}{(s-{1\over \sqrt{2}})^2+{1\over 2}}\,-\,\int \frac{s \sqrt{2}+1}{(s+{1\over \sqrt{2}})^2+{1\over 2}})\,{\rm ds}[/tex]
[tex]I=\ln|\frac{\tan(x)-1}{\tan(x)+1}|\,+\,{1\over 2\sqrt{2}}\ln(-(s-{1\over \sqrt{2}})^2-{1\over 2})\,-\,{1\over 2\sqrt{2}}\ln((s+{1\over \sqrt{2}})^2+{1\over 2})\,+\,C[/tex]
[tex]I=\text 2arctanh(tan(x))\,+\,{1\over 2\sqrt{2}}\ln|\frac{-(\tan(x)-{1\over \sqrt{2}})^2-{1\over 2}}{(\tan(x)+{1\over \sqrt{2}})^2+{1\over 2}}|\,+\,C[/tex]
