ja zrobic taka calke?
\(\displaystyle{ \int\frac{1+ cosx}{1+9sin^2x}\(dx}\)
dzieki
pomysl mój na to to J= \(\displaystyle{ \int\frac{1}{1+9sin^2x}\(dx}\) + \(\displaystyle{ \int\frac{cosx}{1+9sin^2x}\(dx}\)
no i teraz co z ta pierwsza calka?-bo druga wiem.
[ Dodano: 10 Września 2007, 16:16 ]
czy to sie wogole da zrobic?
calka -co dalej?
-
maciej-pop
- Użytkownik
- Posty: 10
- Rejestracja: 24 sie 2007, o 00:00
- Płeć: Mężczyzna
- Lokalizacja: poznan
-
jasny
- Użytkownik
- Posty: 845
- Rejestracja: 2 kwie 2006, o 23:32
- Płeć: Mężczyzna
- Lokalizacja: Limanowa
- Pomógł: 191 razy
calka -co dalej?
Można np podstawienie zrobić \(\displaystyle{ u=\tan\frac{x}{2}}\), wtedy:
\(\displaystyle{ dx=\frac{2du}{1+u^2},\;\sin{x}=\frac{2u}{1+u^2},\;\cos{x}=\frac{1-u^2}{1+u^2}}\)
\(\displaystyle{ I=\int\frac{1+\frac{1-u^2}{1+u^2}}{1+\frac{36u^2}{(1+u^2)^2}}\cdot\frac{2du}{1+u^2}= t\frac{\frac{4du}{(1+u^2)^2}}{\frac{u^4+38u^2+1}{(1+u^2)^2}}=\int\frac{4du}{u^4+38u^2+1}= t\frac{4du}{(u^2+19+6\sqrt{10})(u^2+19-6\sqrt{10})}= -\frac{\sqrt{10}}{30}\int\frac{du}{u^2+19+6\sqrt{10}}+\frac{\sqrt{10}}{30}\int\frac{du}{u^2+19-6\sqrt{10}}}\)
\(\displaystyle{ u=\sqrt{19+6\sqrt{10}}t=\sqrt{(3+\sqrt{10})^2}t=(3+\sqrt{10})t,\;du=(3+\sqrt{10})dt}\)
\(\displaystyle{ u=\sqrt{19-6\sqrt{10}}v=\sqrt{(3-\sqrt{10})^2}v=(\sqrt{10}-3)v,\;du=(\sqrt{10}-3)dv}\)
\(\displaystyle{ I=-\frac{\sqrt{10}}{30}\int\frac{(3+\sqrt{10})dt}{(19+6\sqrt{10})t^2+19+6\sqrt{10}}+\frac{\sqrt{10}}{30}\int\frac{(\sqrt{10}-3)dv}{(19-6\sqrt{10})v^2+19-6\sqrt{10}}= -\frac{\sqrt{10}(3+\sqrt{10})}{30(19+6\sqrt{10})}\int\frac{dt}{t^2+1}+\frac{\sqrt{10}(\sqrt{10}-3)}{30(19-6\sqrt{10})}\int\frac{dv}{v^2+1}=}\) \(\displaystyle{ -\frac{(3\sqrt{10}+10)(19-6\sqrt{10})}{30(19+6\sqrt{10})(19-6\sqrt{10})}arc\tan{t}+\frac{(10-3\sqrt{10})(19+6\sqrt{10})}{30(19-6\sqrt{10})(19+6\sqrt{10})}arc\tan{v}= -\frac{57\sqrt{10}-180+190-60\sqrt{10}}{30}arc\tan\frac{u}{3+\sqrt{10}}+\frac{190+60\sqrt{10}-57\sqrt{10}-180}{30}arc\tan\frac{u}{\sqrt{10}-3}=\frac{3\sqrt{10}-10}{30}arc\tan(\tan\frac{x}{2}(\sqrt{10}-3))+\frac{3\sqrt{10}+10}{30}arc\tan(\tan\frac{x}{2}(\sqrt{10}+3))}\)
\(\displaystyle{ dx=\frac{2du}{1+u^2},\;\sin{x}=\frac{2u}{1+u^2},\;\cos{x}=\frac{1-u^2}{1+u^2}}\)
\(\displaystyle{ I=\int\frac{1+\frac{1-u^2}{1+u^2}}{1+\frac{36u^2}{(1+u^2)^2}}\cdot\frac{2du}{1+u^2}= t\frac{\frac{4du}{(1+u^2)^2}}{\frac{u^4+38u^2+1}{(1+u^2)^2}}=\int\frac{4du}{u^4+38u^2+1}= t\frac{4du}{(u^2+19+6\sqrt{10})(u^2+19-6\sqrt{10})}= -\frac{\sqrt{10}}{30}\int\frac{du}{u^2+19+6\sqrt{10}}+\frac{\sqrt{10}}{30}\int\frac{du}{u^2+19-6\sqrt{10}}}\)
\(\displaystyle{ u=\sqrt{19+6\sqrt{10}}t=\sqrt{(3+\sqrt{10})^2}t=(3+\sqrt{10})t,\;du=(3+\sqrt{10})dt}\)
\(\displaystyle{ u=\sqrt{19-6\sqrt{10}}v=\sqrt{(3-\sqrt{10})^2}v=(\sqrt{10}-3)v,\;du=(\sqrt{10}-3)dv}\)
\(\displaystyle{ I=-\frac{\sqrt{10}}{30}\int\frac{(3+\sqrt{10})dt}{(19+6\sqrt{10})t^2+19+6\sqrt{10}}+\frac{\sqrt{10}}{30}\int\frac{(\sqrt{10}-3)dv}{(19-6\sqrt{10})v^2+19-6\sqrt{10}}= -\frac{\sqrt{10}(3+\sqrt{10})}{30(19+6\sqrt{10})}\int\frac{dt}{t^2+1}+\frac{\sqrt{10}(\sqrt{10}-3)}{30(19-6\sqrt{10})}\int\frac{dv}{v^2+1}=}\) \(\displaystyle{ -\frac{(3\sqrt{10}+10)(19-6\sqrt{10})}{30(19+6\sqrt{10})(19-6\sqrt{10})}arc\tan{t}+\frac{(10-3\sqrt{10})(19+6\sqrt{10})}{30(19-6\sqrt{10})(19+6\sqrt{10})}arc\tan{v}= -\frac{57\sqrt{10}-180+190-60\sqrt{10}}{30}arc\tan\frac{u}{3+\sqrt{10}}+\frac{190+60\sqrt{10}-57\sqrt{10}-180}{30}arc\tan\frac{u}{\sqrt{10}-3}=\frac{3\sqrt{10}-10}{30}arc\tan(\tan\frac{x}{2}(\sqrt{10}-3))+\frac{3\sqrt{10}+10}{30}arc\tan(\tan\frac{x}{2}(\sqrt{10}+3))}\)