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Tính tích phân:
$I = \int\limits_0^{\frac{\pi }{2}} {\sqrt {1 - \sqrt 3 \sin 2x + 2{{\cos }^2}x} dx}$
Bài giải:
Ta có:
$1 - \sqrt 3 \sin 2x + 2{\cos ^2}x$
$= {\sin ^2}x - 2\sqrt 3 \sin x\cos x + 3{\cos ^2}x$
$= {\left( {\sin x - \sqrt 3 \cos x} \right)^2}$
$= {\left[ {2\left( {\frac{1}{2}{\mathop{\rm s}\nolimits} {\rm{inx}} - \frac{{\sqrt 3 }}{2}c{\rm{os}}x} \right)} \right]^2}$
$= {\left[ {2\sin \left( {x - \frac{\pi }{3}} \right)} \right]^2}$
Suy ra:
$\sqrt {1 - \sqrt 3 \sin 2x + 2{{\cos }^2}x}$
$= \sqrt {{{\left[ {2\sin \left( {x - \frac{\pi }{3}} \right)} \right]}^2}}$
$= 2\left| {\sin \left( {x - \frac{\pi }{3}} \right)} \right|$
Do đó:
$I = \int\limits_0^{\frac{\pi }{2}} {\sqrt {1 - \sqrt 3 \sin 2x + 2{{\cos }^2}x} dx}$
$= - 2\int\limits_0^{\frac{\pi }{3}} {\sin \left( {x - \frac{\pi }{3}} \right)dx} + 2\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\sin \left( {x - \frac{\pi }{3}} \right)dx}$
$= 2\left. {c{\rm{os}}\left( {x - \frac{\pi }{3}} \right)} \right|_0^{\frac{\pi }{3}} - 2\left. {c{\rm{os}}\left( {x - \frac{\pi }{3}} \right)} \right|_{\frac{\pi }{3}}^{\frac{\pi }{2}}$
$= 2\left( {1 - \frac{1}{2}} \right) - 2\left( {\frac{{\sqrt 3 }}{2} - 1} \right)$
$= 3 - \sqrt 3$
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