taka całka:
\(\displaystyle{ \int\frac{3-x}{\sqrt{4x - x^2}}dx}\)
i taka:
\(\displaystyle{ \int\frac{e^x( 1 + e^x)}{\sqrt{1 - e^{2x} }}dx}\)
całki 2
- przemk20
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- Rejestracja: 6 gru 2006, o 22:47
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całki 2
1
\(\displaystyle{ \int \frac{3-x}{ \sqrt{4x - x^2 }} = \frac{1}{2} t \frac{-2x+4}{\sqrt{4x-x^2}}dx + t \frac{dx}{4x-x^2} \\
1 \ | 4x-x^2 = t, \ (-2x+4) dx = dt | \\
2 \ | 4x - x^2 = 4 - (x-2)^2 \ | (x-2) = 2t, \ \ 2dt = dx | \\
t \frac{dt}{\sqrt{1-t^2}} = \arcsin t+C = \arcsin (\frac{x-2}{2}) + C \\}\)
2
\(\displaystyle{ e^x = t, \ \ e^x dx = dt \\
t \frac{1+t}{\sqrt{1-t^2}} = \frac{1}{2} t \frac{2t dt}{\sqrt{1-t^2}} + t \frac{1}{\sqrt{1-t^2}}}\)
\(\displaystyle{ \int \frac{3-x}{ \sqrt{4x - x^2 }} = \frac{1}{2} t \frac{-2x+4}{\sqrt{4x-x^2}}dx + t \frac{dx}{4x-x^2} \\
1 \ | 4x-x^2 = t, \ (-2x+4) dx = dt | \\
2 \ | 4x - x^2 = 4 - (x-2)^2 \ | (x-2) = 2t, \ \ 2dt = dx | \\
t \frac{dt}{\sqrt{1-t^2}} = \arcsin t+C = \arcsin (\frac{x-2}{2}) + C \\}\)
2
\(\displaystyle{ e^x = t, \ \ e^x dx = dt \\
t \frac{1+t}{\sqrt{1-t^2}} = \frac{1}{2} t \frac{2t dt}{\sqrt{1-t^2}} + t \frac{1}{\sqrt{1-t^2}}}\)