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Δημοσιεύτηκε: **Κυρ Φεβ 19, 2012 9:54 pm**

$\displaystyle\int_{0}^{\frac{\pi}{4}}\frac{\sin^{2}x\cos^{2}x}{\sin^{3}x+\cos^{3}x}dx$

Απάντηση

Δημοσιεύτηκε: **Παρ Φεβ 24, 2012 11:00 pm**

Κάπως συνοπτικά.

$\displaystyle{I = \int\limits_0^{\pi /4} {\frac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^3}x + {{\cos }^3}x}}dx} = \frac{1}{4}\int\limits_0^{\pi /4} {\frac{{{{\sin }^2}2x}}{{\left( {\sin x + \cos x} \right)\left( {1 - \sin x\cos x} \right)}}dx} = \frac{1}{4}\int\limits_0^{\pi /4} {\frac{{{{\sin }^2}2x}}{{\sqrt {1 + \sin 2x} \left( {1 - \frac{1}{2}\sin 2x} \right)}}dx} = \mathop = \limits^{2x = y} = }$

$\displaystyle{ = \frac{1}{8}\int\limits_0^{\pi /2} {\frac{{{{\sin }^2}y}}{{\sqrt {1 + \sin y} \left( {1 - \frac{1}{2}\sin y} \right)}}dy} = \mathop = \limits^{\sin y = x} = \frac{1}{8}\int\limits_0^1 {\frac{{{x^2}}}{{\left( {1 + x} \right)\left( {1 - \frac{1}{2}x} \right)\sqrt {1 - x} }}dx} = \mathop = \limits^{\sqrt {1 - x} = y} = \frac{1}{2}\int\limits_0^1 {\frac{{{{\left( {1 - {y^2}} \right)}^2}}}{{\left( {2 - {y^2}} \right)\left( {1 + {y^2}} \right)}}dy} = }$

$\displaystyle{ = \frac{1}{2}\int\limits_0^1 {\frac{{{y^2}\left( {{y^2} - 2} \right) + 1}}{{\left( {2 - {y^2}} \right)\left( {1 + {y^2}} \right)}}dy} = \frac{1}{2}\int\limits_0^1 {\left( { - \frac{{{y^2}}}{{1 + {y^2}}} + \frac{1}{{\left( {2 - {y^2}} \right)\left( {1 + {y^2}} \right)}}} \right)dy} = \frac{1}{2}\int\limits_0^1 {\left( { - 1 + \frac{4}{3} \cdot \frac{1}{{1 + {y^2}}} + \frac{1}{3}\frac{1}{{2 - {y^2}}}} \right)dy} = }$

$\displaystyle{ = - \frac{1}{2} + \frac{\pi }{6} - \frac{1}{{12\sqrt 2 }}\int\limits_0^1 {\left( {\frac{1}{{y - \sqrt 2 }} - \frac{1}{{y + \sqrt 2 }}} \right)dy} = \frac{{\pi - 3}}{6} - \frac{{\sqrt 2 \ln \left( {\sqrt 2 - 1} \right)}}{{12}}}$

Δημοσιεύτηκε: **Παρ Φεβ 24, 2012 11:14 pm**

Έυχαριστώ Σεραφείμ , η λύση που έχω δει http://www.artofproblemsolving.com/Foru ... 6&t=463353

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