Rozwiązanie:
Zadanie [tex]\bold{1.}[/tex]
Krzywa:
[tex]y=\ln \sin x[/tex]
Długość łuku:
[tex]$|L|=\int\limits^b_a {\sqrt{1+(y')^{2}} \, dx[/tex]
Mamy:
[tex]$y'=\frac{1}{\sin x} \cdot (\sin x)'=\frac{\cos x}{\sin x} =\cot x[/tex]
[tex]$|L|=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {\sqrt{1+\cot^{2} x } \, dx =\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {\sqrt{1+\frac{\cos^{2} x}{\sin^{2} x} } \, dx=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {\sqrt{\frac{\sin ^{2}x+\cos^{2}x}{\sin^{2}x} } \, dx=[/tex]
[tex]$=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {\sqrt{\frac{1}{\sin^{2}x} }} \, dx=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} \frac{1}{\sin x} \, dx[/tex]
Całka nieoznaczona (liczymy podstawieniem uniwersalnym):
[tex]$\int {\frac{1}{\sin x} } \, dx=\left|\begin{array}{ccc}u=\tan \frac{x}{2} \\\frac{x}{2}=\arctan u\\dx=\frac{2}{1+u^{2}} \ du \end{array}\right|=\int {\frac{1}{\frac{2u}{1+u^{2}}} } \, \cdot \frac{2}{1+u^{2}} \ du=\int {\frac{1+u^{2}}{2u} } \, \cdot \frac{2}{1+u^{2}} \ du=[/tex]
[tex]$=\int {\frac{1}{u} } \, du=\ln u+C=\ln \tan \frac{x}{2} +C[/tex]
Zatem:
[tex]$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} \frac{1}{\sin x} \, dx=\ln \tan \frac{x}{2} \Big|^{\frac{\pi}{2}}_{\frac{\pi}{3}}=\ln \tan \frac{\pi}{4} -\ln \tan \frac{\pi}{6}=\ln 1-\ln \frac{\sqrt{3}}{3}=-\ln \frac{\sqrt{3}}{3}=\frac{\ln3}{2}[/tex]
Zadanie [tex]\bold{2.}[/tex]
Krzywe:
[tex]y=3\sin x[/tex]
[tex]y=\sin x[/tex]
Objętość:
[tex]$|V|=\pi \int\limits^b_a y^{2}\ dx[/tex]
Mamy:
[tex]$|V|=\pi \int\limits^{\pi}_{0} {9\sin^{2} x-\sin^{2} x} \, dx=8\pi \int\limits^{\pi}_{0}\sin^{2}x \ dx=8\pi \cdot \frac{2x-\sin 2x}{4} \Bigg|^{\pi}_{0}=[/tex]
[tex]$=2\pi (2x-\sin 2x)\Big|^{\pi}_{0}=4\pi^{2}[/tex]
