Obliczyć długość łuku krzywej \(y = 2ln(cosx)\) ; \(0 < x < \frac{\pi}{4}\)
Dostałem całkę \(\int \sqrt{1 + 4 (\tg x)^{2}} dx\) i nie wiem co dalej z nią zrobić. Proszę o pomoc.
Długość łuku krzywej
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Robakks
- Czasem tu bywam
- Posty: 149
- Rejestracja: 30 wrz 2012, 20:36
- Podziękowania: 2 razy
- Otrzymane podziękowania: 13 razy
- Płeć:
\(\int \sqrt{1 + 4 (\tan^{2} {x})} dx\\
\sqrt{1 + 4 (\tan^{2}{x})}=t-2\tan{x}\\
1 + 4 \tan^{2}{x}=t^2-4t\tan{x}+4\tan^{2}{x}\\
1=t^2-4t\tan{x}\\
4t\tan{x}=t^2-1\\
\tan{x}=\frac{t^2-1}{4t}\\
t-2\tan{x}=\frac{4t^2-2t^2+2}{4t}=\frac{t^2+1}{2t}\\
\frac{\cos{x} \cdot \cos{x}- \left(-\sin{x} \right)\sin{x} }{\cos^{2}{x}}dx=\frac{2t \cdot 4t-4 \left(t^2-1 \right) }{16t^2}dt\\
\left(1+\tan^{2}{x} \right)dx=\frac{4t^2+4}{16t^2} \\
\left(\frac{16t^2+ \left( t^2-1\right)^2 }{16t^2} \right)dx=\frac{t^2+1}{4t^2} dt\\
dx=\frac{t^2+1}{4t^2} \cdot \frac{16t^2}{t^4+14t^2+1}dt\\
dx=4\frac{t^2+1}{t^4+14t^2+1}dt\\
4\int{\frac{t^2+1}{2t} \cdot \frac{t^2+1}{t^4+14t^2+1}dt}\\
2\int{\frac{ \left(t^2+1 \right)^2 }{t \left( t^4+14t^2+1\right) }dt}\\
2\int{\frac{ t\left(t^2+1 \right)^2 }{t^2 \left( t^4+14t^2+1\right) }dt}\\
u=t^2\\
du=2tdt\\
\int{\frac{ \left( u+1\right)^2 }{u \left(u^2+14u+1 \right) }du}\\
\int{\frac{ u^2+2u+1 }{u \left(u^2+14u+1 \right) }du}\\
\int{\frac{u^2+14u+1-12u}{u \left(u^2+14u+1 \right)}du}\\
\int{\frac{du}{u}}-12\int{\frac{du}{u^2+14u+1}}\\
\int{\frac{du}{u}}-12\int{\frac{du}{\left(u+7-4 \sqrt{3} \right)\left(u+7+4 \sqrt{3} \right)}}\\
\int{\frac{du}{u}}+\frac{ \sqrt{3} }{2}\int{\frac{\left(u+7-4 \sqrt{3} \right)-\left(u+7+4 \sqrt{3} \right)}{\left(u+7-4 \sqrt{3} \right)\left(u+7+4 \sqrt{3} \right)}du}\\
\ln{\left|u\right|}+\frac{ \sqrt{3} }{2}\ln{\left|\frac{u+7+4 \sqrt{3}}{u+7-4 \sqrt{3}}\right|}+C\\\)
\sqrt{1 + 4 (\tan^{2}{x})}=t-2\tan{x}\\
1 + 4 \tan^{2}{x}=t^2-4t\tan{x}+4\tan^{2}{x}\\
1=t^2-4t\tan{x}\\
4t\tan{x}=t^2-1\\
\tan{x}=\frac{t^2-1}{4t}\\
t-2\tan{x}=\frac{4t^2-2t^2+2}{4t}=\frac{t^2+1}{2t}\\
\frac{\cos{x} \cdot \cos{x}- \left(-\sin{x} \right)\sin{x} }{\cos^{2}{x}}dx=\frac{2t \cdot 4t-4 \left(t^2-1 \right) }{16t^2}dt\\
\left(1+\tan^{2}{x} \right)dx=\frac{4t^2+4}{16t^2} \\
\left(\frac{16t^2+ \left( t^2-1\right)^2 }{16t^2} \right)dx=\frac{t^2+1}{4t^2} dt\\
dx=\frac{t^2+1}{4t^2} \cdot \frac{16t^2}{t^4+14t^2+1}dt\\
dx=4\frac{t^2+1}{t^4+14t^2+1}dt\\
4\int{\frac{t^2+1}{2t} \cdot \frac{t^2+1}{t^4+14t^2+1}dt}\\
2\int{\frac{ \left(t^2+1 \right)^2 }{t \left( t^4+14t^2+1\right) }dt}\\
2\int{\frac{ t\left(t^2+1 \right)^2 }{t^2 \left( t^4+14t^2+1\right) }dt}\\
u=t^2\\
du=2tdt\\
\int{\frac{ \left( u+1\right)^2 }{u \left(u^2+14u+1 \right) }du}\\
\int{\frac{ u^2+2u+1 }{u \left(u^2+14u+1 \right) }du}\\
\int{\frac{u^2+14u+1-12u}{u \left(u^2+14u+1 \right)}du}\\
\int{\frac{du}{u}}-12\int{\frac{du}{u^2+14u+1}}\\
\int{\frac{du}{u}}-12\int{\frac{du}{\left(u+7-4 \sqrt{3} \right)\left(u+7+4 \sqrt{3} \right)}}\\
\int{\frac{du}{u}}+\frac{ \sqrt{3} }{2}\int{\frac{\left(u+7-4 \sqrt{3} \right)-\left(u+7+4 \sqrt{3} \right)}{\left(u+7-4 \sqrt{3} \right)\left(u+7+4 \sqrt{3} \right)}du}\\
\ln{\left|u\right|}+\frac{ \sqrt{3} }{2}\ln{\left|\frac{u+7+4 \sqrt{3}}{u+7-4 \sqrt{3}}\right|}+C\\\)