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proszę o pomoc, tym razem funkcja wygląda tak: y=3-ln(sinx) (pi/6;pi/3)

-- 10 lut 2009, o 16:15 --

oraz całka:
\(\displaystyle{ \int_{}^{}}\) \(\displaystyle{ \frac{5x}{ 2x^{2} -2x+1 }}\) dx

\(\displaystyle{ y'=-\cot x}\)

\(\displaystyle{ (y')^{2}=\cot ^{2} x}\)

\(\displaystyle{ |\Gamma |=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1+\frac{\cos ^{2}x}{\sin ^{2}x}} dx = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\sin x} =...}\)

mlaczyn pisze: \(\displaystyle{ \int \frac{5x}{ 2x^{2} -2x+1 }dx}\)

\(\displaystyle{ ...= \int \frac{ \frac{5}{4}(2x ^{2}-2x+1)'+ \frac{5}{2} }{2x ^{2}-2x+1 }= \frac{5}{4} ln|2x ^{2}-2x+1|+ \frac{5}{4} \int \frac{dx}{(x- \frac{1}{2}) ^{2}+ (\frac{1}{2}) ^{2} }=}\)

\(\displaystyle{ \frac{5}{4} ln|2x ^{2}-2x+1|+ \frac{5}{2}arctg(2x-1)+C}\)

znowu łuk;/ t (90;135)

x(t)=\(\displaystyle{ cos^{3}}\)2t

y(t)=\(\displaystyle{ sin ^{3}}\)2t