Obliczyć całkę oznaczoną.... Question from @Martynajaaaa

January 2019 1 19 Report

Obliczyć całkę oznaczoną

Isoko F(x) - funkcja pierwotna

$\displaystyle \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\cos x}= F\left(\frac{\pi}{3}\right)-F\left(\frac{\pi}{6}\right)$

Tę całkę liczymy przez podstawienie:

$t= \hbox{tg} \ \frac{x}{2} \\ \\ dx=\dfrac{2}{1+t^2}dt \\ \\ \cos x = \dfrac{1-t^2}{1+t^2} \\ \\ \displaystyle \int \frac{dx}{\cos x}=\int \frac{\frac{2}{1+t^2}dt}{\frac{1-t^2}{1+t^2}}=\int \frac{2}{1-t^2}dt=2\int \frac{dt}{1-t^2}= \\ \\ =2 \cdot \frac{1}{2} \ln \left| \frac{1+t}{1-t}\right|+C=\ln \left| \frac{1+t}{1-t}\right|+C$

Wzór:

$\displaystyle \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \left| \frac{a+x}{a-x}\right|+C$

Powróć do podstawienia:

$F(x) = \ln \left| \dfrac{1+\hbox{tg}\frac{x}{2}}{1-\hbox{tg} \frac{x}{2}}\right|$

Warto wiedzieć ile wynosi tg(pi/12). Ze wzoru $\hbox{tg} \frac{x}{2}=\dfrac{\sin x}{1+\cos x}$ otrzymamy, że

$\hbox{tg} \ \frac{\pi}{12}= \dfrac{\sin \frac{\pi}{6}}{1+\cos \frac{\pi}{6}} = \dfrac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}=\dfrac{1}{2+\sqrt{3}}=\dfrac{2-\sqrt{3}}{1}=2-\sqrt{3}$

$F\left( \dfrac{\pi}{6}\right)=\ln \left| \dfrac{1+\hbox{tg} \frac{\pi}{12}}{1-\hbox{tg} \frac{\pi}{12}}\right| = \ln \left| \dfrac{1+2-\sqrt{3}}{1-2+\sqrt{3}}\right|=\ln \left| \dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right|= \\ \\ = \ln \left| \dfrac{(3-\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\right|=\ln \left| \dfrac{2\sqrt{3}}{2}\right|=\ln |\sqrt{3}|=\ln \sqrt{3}$

$F\left( \dfrac{\pi}{3}\right)= \ln \left| \dfrac{1+\hbox{tg} \ \frac{\pi}{6}}{1-\hbox{tg} \frac{\pi}{6}}\right|= \ln \left| \dfrac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}\right|=\ln \left| \dfrac{3+\sqrt{3}}{3-\sqrt{3}}\right|= \\ \\ = \ln \left| \dfrac{(3+\sqrt{3})^2}{(3-\sqrt{3})(3+\sqrt{3})}\right|=\ln \left| \dfrac{12+6\sqrt{3}}{6}\right|=\ln (2+\sqrt{3})$

Podstawiwam do wzoru na samym początku

$\displaystyle \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\cos x}= \ln (2+\sqrt{3})-\ln(\sqrt{3})= \ln \dfrac{2+\sqrt{3}}{\sqrt{3}}=\ln \left(\dfrac{2\sqrt{3}+3}{3}\right)$

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