Bài 104354

Question

Tính tích phân $I=\int\limits_{0}^{\pi/2}\frac{sinx+sin2x}{\sqrt{3sin^2x+4cos^2x}}dx$

hoàng anh thọ · Answer 1 · 5/16/2012 10:44:22 AM

$I = \int\limits_0^{\pi /2} {\frac{{\sin x + \sin 2x}}{{\sqrt {3{{\sin }^2}x + 4co{s^2}x} }}} dx = \int\limits_0^{\pi /2} {\frac{{\sin x}}{{\sqrt {3{{\sin }^2}x + 4co{s^2}x} }}} dx + \int\limits_0^{\pi /2} {\frac{{\sin 2x}}{{\sqrt {3{{\sin }^2}x + 4co{s^2}x} }}} dx$ $=A+B$
Có $A = \int\limits_0^{\pi /2} {\frac{{\sin x}}{{\sqrt {3{{\sin }^2}x + 4co{s^2}x} }}} dx = - \int\limits_0^{\pi /2} {\frac{{d\left( {\cos x} \right)}}{{\sqrt {3 + co{s^2}x} }}} \underbrace { = }_{ t=\cos { x } } \int\limits_0^1 {\frac{{dt}}{{\sqrt {3 + {t^2}} }}} $
Đặt $t = \sqrt 3 \tan u \Rightarrow dt = \sqrt 3.\frac{1}{\cos^2 u}du= \sqrt 3 \left( {1 + {{\tan }^2}u} \right)du$ thì
$\Rightarrow A = \int\limits_0^{\pi /6} {\frac{{\sqrt 3 \left( {1 + {{\tan }^2}u} \right)du}}{{\sqrt {3 + 3{{\tan }^2}u} }}} = \int\limits_0^{\pi /6} {\frac{{du}}{{\cos u}}} = \int\limits_0^{\pi /6} {\frac{{d\left( {\sin u} \right)}}{{1 - {{\sin }^2}u}}} = \frac{1}{2}\left. {\ln \left| {\frac{{1 + \sin u}}{{1 - \sin u}}} \right|} \right|_0^{\pi /6} = \frac{1}{2}\ln 3$
$B = \int\limits_0^{\pi /2} {\frac{{\sin 2x}}{{\sqrt {3{{\sin }^2}x + 4co{s^2}x} }}} dx = - \int\limits_0^{\pi /2} {\frac{{d\left( {4 - {{\sin }^2}x} \right)}}{{\sqrt {4 - {{\sin }^2}x} }}} = - \left. {2\sqrt {4 - {{\sin }^2}x} } \right|_0^{\pi /2} = 2\left( {2 - \sqrt 3 } \right)$
Vậy $I = A + B = \boxed{\frac{{\ln 3}}{2} + 2\left( {2 - \sqrt 3 } \right)}$

Bài 104354

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Bài 104354

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