Ολοκληρωματάκι

#1

$\displaystyle\int_{0}^{\pi}\frac{x}{1+e\sin x}\,dx$ .

Ας δούμε μετά και αυτό :

$\displaystyle\int\frac{x}{1+\sin x}\,dx$ .

Δεν το χω λύσει...

mathxl · #2

Για το αόριστο, μπορούμε και σχολικά
$\displaystyle{\begin{array}{l} \displaystyle \int \frac{x}{{1 + \sin x}}\,dx = \int {\frac{x}{{1 + \cos \left( {\frac{\pi }{2} - x} \right)}}} dx = \int {\frac{x}{{2{{\cos }^2}\left( {\frac{\pi }{4} - \frac{x}{2}} \right)}}} dx = \\ =\displaystyle\ - x\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right) + \int {\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right)dx = } \\ =\displaystyle\ - x\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right) + 2\ln \left| {\cos \left( {\frac{\pi }{4} - \frac{x}{2}} \right)} \right| + c \\ \end{array}}$

mathxl · #3

Το ορισμένο μου βγήκε λίγο μεγάλο
$\displaystyle{\begin{array}{l} I = \int\limits_0^\pi {\frac{{xdx}}{{1 + e\sin x}}} \mathop = \limits_{dx = - du}^{x = \pi - u} \int\limits_0^\pi {\frac{{\pi - u}}{{1 + e\sin u}}du} = \int\limits_0^\pi {\frac{\pi }{{1 + e\sin u}}du} - I \Rightarrow \\ \displaystyle\ 2I = \pi \int\limits_0^\pi {\frac{1}{{1 + e\sin u}}du} \mathop = \limits_{du = 2dt}^{u = 2t} 2\pi \int\limits_0^{\frac{\pi }{2}} {\frac{1}{{1 + e\sin 2t}}dt} \Leftrightarrow \\ I = \pi \int\limits_0^{\frac{\pi }{2}} {\frac{1}{{1 + \frac{{2e\tan t}}{{1 + {{\tan }^2}t}}}}dt} = \pi \int\limits_0^{\frac{\pi }{2}} {\frac{{1 + {{\tan }^2}t}}{{1 + {{\tan }^2}t + 2e\tan t}}dt} = \\ \end{array}}$
$\displaystyle{\begin{array}{l} = \pi \int\limits_0^{\frac{\pi }{2}} {\frac{{1 + {{\tan }^2}t}}{{1 + {{\tan }^2}t + 2e\tan t + {e^2} - {e^2}}}dt} = \\ = \pi \int\limits_0^{\frac{\pi }{2}} {\frac{{1 + {{\tan }^2}t}}{{{{\left( {\tan t + e} \right)}^2} + 1 - {e^2}}}dt} \mathop = \limits_{\left( {1 + {{\tan }^2}t} \right)dt = dy}^{\tan t + e = y} \\ = \pi \int\limits_e^{ + \infty } {\frac{1}{{{y^2} + 1 - {e^2}}}dy} = \pi \int\limits_e^{ + \infty } {\frac{1}{{\left( {y - \sqrt {{e^2} - 1} } \right)\left( {y + \sqrt {{e^2} - 1} } \right)}}dy} = \\ \end{array}}$
$\displaystyle{\begin{array}{l} = - \frac{\pi }{{2\sqrt {{e^2} - 1} }}\int\limits_e^{ + \infty } {\frac{{y - \sqrt {{e^2} - 1} - y - \sqrt {{e^2} - 1} }}{{\left( {y - \sqrt {{e^2} - 1} } \right)\left( {y + \sqrt {{e^2} - 1} } \right)}}dy} = \\ = - \frac{\pi }{{2\sqrt {{e^2} - 1} }}\int\limits_e^{ + \infty } {\left( {\frac{1}{{y + \sqrt {{e^2} - 1} }} - \frac{1}{{y - \sqrt {{e^2} - 1} }}} \right)} dy = \\ = - \frac{\pi }{{2\sqrt {{e^2} - 1} }}\left[ {\ln \left| {\frac{{y + \sqrt {{e^2} - 1} }}{{y - \sqrt {{e^2} - 1} }}} \right|} \right]_e^{ + \infty } = \\ \end{array}}$
$\displaystyle{\begin{array}{l} = - \frac{\pi }{{2\sqrt {{e^2} - 1} }}\left( {0 - \ln \left| {\frac{{e + \sqrt {{e^2} - 1} }}{{e - \sqrt {{e^2} - 1} }}} \right|} \right) = \\ = \frac{\pi }{{2\sqrt {{e^2} - 1} }}\ln \left( {\frac{{e + \sqrt {{e^2} - 1} }}{{e - \sqrt {{e^2} - 1} }}} \right) \\ \end{array}}$

#4

mathxl έγραψε:Το ορισμένο μου βγήκε λίγο μεγάλο
$\displaystyle{\begin{array}{l} I = \int\limits_0^\pi {\frac{{xdx}}{{1 + e\sin x}}} \mathop = \limits_{dx = - du}^{x = \pi - u} \int\limits_0^\pi {\frac{{\pi - u}}{{1 + e\sin u}}du} = \int\limits_0^\pi {\frac{\pi }{{1 + e\sin u}}du} - I \Rightarrow \\ \displaystyle\ 2I = \pi \int\limits_0^\pi {\frac{1}{{1 + e\sin u}}du} \mathop = \limits_{du = 2dt}^{u = 2t} 2\pi \int\limits_0^{\frac{\pi }{2}} {\frac{1}{{1 + e\sin 2t}}dt} \Leftrightarrow \\ I = \pi \int\limits_0^{\frac{\pi }{2}} {\frac{1}{{1 + \frac{{2e\tan t}}{{1 + {{\tan }^2}t}}}}dt} = \pi \int\limits_0^{\frac{\pi }{2}} {\frac{{1 + {{\tan }^2}t}}{{1 + {{\tan }^2}t + 2e\tan t}}dt} = \\ \end{array}}$
$\displaystyle{\begin{array}{l} = \pi \int\limits_0^{\frac{\pi }{2}} {\frac{{1 + {{\tan }^2}t}}{{1 + {{\tan }^2}t + 2e\tan t + {e^2} - {e^2}}}dt} = \\ = \pi \int\limits_0^{\frac{\pi }{2}} {\frac{{1 + {{\tan }^2}t}}{{{{\left( {\tan t + e} \right)}^2} + 1 - {e^2}}}dt} \mathop = \limits_{\left( {1 + {{\tan }^2}t} \right)dt = dy}^{\tan t + e = y} \\ = \pi \int\limits_e^{ + \infty } {\frac{1}{{{y^2} + 1 - {e^2}}}dy} = \pi \int\limits_e^{ + \infty } {\frac{1}{{\left( {y - \sqrt {{e^2} - 1} } \right)\left( {y + \sqrt {{e^2} - 1} } \right)}}dy} = \\ \end{array}}$
$\displaystyle{\begin{array}{l} = - \frac{\pi }{{2\sqrt {{e^2} - 1} }}\int\limits_e^{ + \infty } {\frac{{y - \sqrt {{e^2} - 1} - y - \sqrt {{e^2} - 1} }}{{\left( {y - \sqrt {{e^2} - 1} } \right)\left( {y + \sqrt {{e^2} - 1} } \right)}}dy} = \\ = - \frac{\pi }{{2\sqrt {{e^2} - 1} }}\int\limits_e^{ + \infty } {\left( {\frac{1}{{y + \sqrt {{e^2} - 1} }} - \frac{1}{{y - \sqrt {{e^2} - 1} }}} \right)} dy = \\ = - \frac{\pi }{{2\sqrt {{e^2} - 1} }}\left[ {\ln \left| {\frac{{y + \sqrt {{e^2} - 1} }}{{y - \sqrt {{e^2} - 1} }}} \right|} \right]_e^{ + \infty } = \\ \end{array}}$
$\displaystyle{\begin{array}{l} = - \frac{\pi }{{2\sqrt {{e^2} - 1} }}\left( {0 - \ln \left| {\frac{{e + \sqrt {{e^2} - 1} }}{{e - \sqrt {{e^2} - 1} }}} \right|} \right) = \\ = \frac{\pi }{{2\sqrt {{e^2} - 1} }}\ln \left( {\frac{{e + \sqrt {{e^2} - 1} }}{{e - \sqrt {{e^2} - 1} }}} \right) \\ \end{array}}$

Ωραίος. Και εγώ κάτι τέτοιο έχω βγάλει με παρόμοιο τρόπο.Το βιβλίο δίνει $\frac{\pi\cos^{-1}e}{\sqrt{1-e^{2}}}$ , που προφανώς είνια λάθος, τουλάχιστον λόγω του ριζικού. Τωρα το τόξο συνημιτόνου δεν ξέρω από πού βγαίνει...

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