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Δημοσιεύτηκε: **Τετ Ιαν 04, 2012 1:40 am**

$\displaystyle\int_{-1}^{1}\frac{x^{3}}{\sqrt{1-x^{2}}}\ln\Big(\frac{1+x}{1-x}\Big)\,dx$

Δημοσιεύτηκε: **Τετ Ιαν 04, 2012 9:13 am**

Υπολογίζουμε το αόριστο ολοκλήρωμα $\displaystyle{I = \int {\frac{{{x^3}}}{{\sqrt {1 - {x^2}} }}dx} }$

$\displaystyle{I = \int {\frac{{{x^3}}}{{\sqrt {1 - {x^2}} }}dx} = \mathop = \limits^{x = \sqrt y } = \frac{1}{2}\int {\frac{y}{{\sqrt {1 - y} }}dy} = \mathop = \limits^{\sqrt {1 - y} = z} = \int {\left( {{z^2} - 1} \right)dz} = \frac{1}{3}{z^3} - z = .. = - \frac{1}{3}\sqrt {1 - {x^2}} \left( {{x^2} + 2} \right)}$ .

Τότε $\displaystyle{\int\limits_{ - 1}^1 {\frac{{{x^3}}}{{\sqrt {1 - {x^2}} }}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)dx} = - \frac{1}{3}\int\limits_{ - 1}^1 {{{\left( {\sqrt {1 - {x^2}} \left( {{x^2} + 2} \right)} \right)}{'}}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)dx} \mathop = \limits^{\left[ * \right]} \frac{2}{3}\int\limits_{ - 1}^1 {\frac{{\sqrt {1 - {x^2}} \left( {{x^2} + 2} \right)}}{{1 - {x^2}}}dx} = }$

$\displaystyle{ = \frac{2}{3}\int\limits_{ - 1}^1 {\frac{{{x^2} + 2}}{{\sqrt {1 - {x^2}} }}dx} = \mathop = \limits^{x = \sin y} = \frac{2}{3}\int\limits_{ - \pi /2}^{\pi /2} {\left( {{{\sin }^2}\left( y \right) + 2} \right)dy} \mathop = \limits^{\left[ {**} \right]} \frac{1}{3}\int\limits_{ - \pi /2}^{\pi /2} {\left( {5 - \cos 2y} \right)dy} = \frac{{5\pi }}{3}}$

$\displaystyle{\left[ * \right]:}$ Τα όρια $\displaystyle{\mathop {\lim }\limits_{x \to - 1} \left[ {\sqrt {1 - {x^2}} \left( {{x^2} + 2} \right)\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)} \right]}$ και $\displaystyle{\mathop {\lim }\limits_{x \to 1} \left[ {\sqrt {1 - {x^2}} \left( {{x^2} + 2} \right)\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)} \right]}$ προκύπτουν (κλασσικά) ίσα με μηδέν και $\displaystyle{{\left( {\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)} \right){'}} = \frac{2}{{1 - {x^2}}}}$

$\displaystyle{\left[ {**} \right]:}$ Φανερά $\displaystyle{\int\limits_{ - \pi /2}^{\pi /2} {\cos 2ydx} = 0}$ .

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μεταμεσονύκτιο ολοκλήρωμα 89

μεταμεσονύκτιο ολοκλήρωμα 89

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