Tích phân I = pi/6 ∫ 0 sin^2 x dx có giá trị là

40/91

Tích phân \[I = \int\limits_0^{\frac{\pi }{6}} {{{\sin }^2}x{\rm{d}}x} \]có giá trị là

\[\frac{\pi }{{12}} + \frac{{\sqrt 3 }}{8}\].

\[\frac{\pi }{{12}} - \frac{{\sqrt 3 }}{8}\].

\[ - \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{8}\].

\[\frac{\pi }{{12}} - \frac{{\sqrt 3 }}{4}\].

Giải thích

Đáp án đúng là: B

\[\begin{array}{l}I = \int\limits_0^{\frac{\pi }{6}} {{{\sin }^2}x{\rm{d}}x} = \int\limits_0^{\frac{\pi }{6}} {\frac{{1 - {\rm{cos}}2x}}{2}{\rm{d}}x} = \frac{1}{2}\int\limits_0^{\frac{\pi }{6}} {\left( {1 - {\rm{cos}}2x} \right){\rm{d}}x} = \frac{1}{2}\left. {\left( {x - \frac{1}{2}\sin 2x} \right)} \right|_0^{\frac{\pi }{6}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {\frac{\pi }{6} - \frac{1}{2}\sin \frac{\pi }{3}} \right) = \frac{\pi }{{12}} - \frac{{\sqrt 3 }}{8}.\end{array}\]