mam taką całkę do policzenia:
\(\displaystyle{ \int_{0}^{pi/4} \sqrt{1+ tg x^{2} } dx}\)
dziękuję
całka do policzenia
-
Mariusz M
- Użytkownik
- Posty: 6953
- Rejestracja: 25 wrz 2007, o 01:03
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- Podziękował: 2 razy
- Pomógł: 1254 razy
całka do policzenia
Masz do dyspozycji podstawienia
I podstawienie Eulera
\(\displaystyle{ t=\tan{x}+ \sqrt{1+\tan^{2}{x}}}\)
II podstawienie Eulera
\(\displaystyle{ \sqrt{1+\tan^{2}{x}}=t\tan{x}+1}\)
Podstawienie
\(\displaystyle{ t=\sin{x}}\)
Do ciebie należy wybór
Jeżeli wybierzesz I podstawienie Eulera to powinnaś otrzymać
\(\displaystyle{ \int_{1}^{1+ \sqrt{2} }{ \frac{1}{t} \mbox{d}t }}\)
\(\displaystyle{ =\ln{ \left| 1+ \sqrt{2} \right| }}\)
\(\displaystyle{ \int_{0}^{pi/4} \sqrt{1+ tg x^{2} } dx}\)
\(\displaystyle{ t=\tan{x}+ \sqrt{1+\tan^{2}{x}}}\)
\(\displaystyle{ t-\tan{x}= \sqrt{1+\tan^{2}{x}}}\)
\(\displaystyle{ t^2-2t\tan{x}+tan^{2}{x}= 1+\tan^{2}{x}}\)
\(\displaystyle{ t^2-2t\tan{x}= 1}\)
\(\displaystyle{ t^2-1=2t\tan{x}}\)
\(\displaystyle{ \tan{x}=\frac{t^2-1}{2t}}\)
\(\displaystyle{ \left(1+\tan^{2}{x} \right) \mbox{d}x =\frac{2t \cdot 2t-2 \left(t^2-1 \right) }{4t^2} \mbox{d}t}\)
\(\displaystyle{ \left(1+\tan^{2}{x} \right) \mbox{d}x =\frac{4t^2-2 t^2+2 }{4t^2} \mbox{d}t}\)
\(\displaystyle{ \left(1+\tan^{2}{x} \right) \mbox{d}x =\frac{2 t^2+2 }{4t^2} \mbox{d}t}\)
\(\displaystyle{ \left(1+\tan^{2}{x} \right) \mbox{d}x =\frac{ t^2+1 }{2t^2} \mbox{d}t}\)
\(\displaystyle{ \sqrt{1+\tan^{2}{x}}=t-\tan{x}=t- \frac{t^2-1}{2t}= \frac{2t^2-t^2+1}{2t}= \frac{t^2+1}{2t}}\)
\(\displaystyle{ t \left(0 \right)=0+ \sqrt{1+0}=1}\)
\(\displaystyle{ t \left( \frac{\pi}{4} \right)=1+ \sqrt{1+1}=1+ \sqrt{2}}\)
\(\displaystyle{ \int_{1}^{1+ \sqrt{2} }{ \frac{2t}{t^2+1} \cdot \frac{t^2+1}{2t^2} \mbox{d}t}}\)
\(\displaystyle{ \int_{1}^{1+ \sqrt{2} }{ \frac{1}{t} \mbox{d}t }}\)
\(\displaystyle{ =\ln{ \left| 1+ \sqrt{2} \right| }}\)
I podstawienie Eulera
\(\displaystyle{ t=\tan{x}+ \sqrt{1+\tan^{2}{x}}}\)
II podstawienie Eulera
\(\displaystyle{ \sqrt{1+\tan^{2}{x}}=t\tan{x}+1}\)
Podstawienie
\(\displaystyle{ t=\sin{x}}\)
Do ciebie należy wybór
Jeżeli wybierzesz I podstawienie Eulera to powinnaś otrzymać
\(\displaystyle{ \int_{1}^{1+ \sqrt{2} }{ \frac{1}{t} \mbox{d}t }}\)
\(\displaystyle{ =\ln{ \left| 1+ \sqrt{2} \right| }}\)
\(\displaystyle{ \int_{0}^{pi/4} \sqrt{1+ tg x^{2} } dx}\)
\(\displaystyle{ t=\tan{x}+ \sqrt{1+\tan^{2}{x}}}\)
\(\displaystyle{ t-\tan{x}= \sqrt{1+\tan^{2}{x}}}\)
\(\displaystyle{ t^2-2t\tan{x}+tan^{2}{x}= 1+\tan^{2}{x}}\)
\(\displaystyle{ t^2-2t\tan{x}= 1}\)
\(\displaystyle{ t^2-1=2t\tan{x}}\)
\(\displaystyle{ \tan{x}=\frac{t^2-1}{2t}}\)
\(\displaystyle{ \left(1+\tan^{2}{x} \right) \mbox{d}x =\frac{2t \cdot 2t-2 \left(t^2-1 \right) }{4t^2} \mbox{d}t}\)
\(\displaystyle{ \left(1+\tan^{2}{x} \right) \mbox{d}x =\frac{4t^2-2 t^2+2 }{4t^2} \mbox{d}t}\)
\(\displaystyle{ \left(1+\tan^{2}{x} \right) \mbox{d}x =\frac{2 t^2+2 }{4t^2} \mbox{d}t}\)
\(\displaystyle{ \left(1+\tan^{2}{x} \right) \mbox{d}x =\frac{ t^2+1 }{2t^2} \mbox{d}t}\)
\(\displaystyle{ \sqrt{1+\tan^{2}{x}}=t-\tan{x}=t- \frac{t^2-1}{2t}= \frac{2t^2-t^2+1}{2t}= \frac{t^2+1}{2t}}\)
\(\displaystyle{ t \left(0 \right)=0+ \sqrt{1+0}=1}\)
\(\displaystyle{ t \left( \frac{\pi}{4} \right)=1+ \sqrt{1+1}=1+ \sqrt{2}}\)
\(\displaystyle{ \int_{1}^{1+ \sqrt{2} }{ \frac{2t}{t^2+1} \cdot \frac{t^2+1}{2t^2} \mbox{d}t}}\)
\(\displaystyle{ \int_{1}^{1+ \sqrt{2} }{ \frac{1}{t} \mbox{d}t }}\)
\(\displaystyle{ =\ln{ \left| 1+ \sqrt{2} \right| }}\)