Tích phân
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
$
Tích phân
Tích phân
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
Tích phân
Tích phân
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
$
Tích phân