Đáp án:
a) $\dfrac{\pi }{6}$
b) $\frac{\pi }{3}$
Giải thích các bước giải:
\(a)\int\limits_{\dfrac{{\sqrt 3 }}{3}}^{\sqrt 3 } {\dfrac{1}{{{x^2} + 1}}dx} \)
Đặt \(x = \tan t\), đổi cận \(x = \dfrac{{\sqrt 3 }}{3} \Rightarrow t = \dfrac{\pi }{6},x = \sqrt 3 \Rightarrow t = \dfrac{\pi }{3}\)
\(x = \tan t \Rightarrow dx = \left( {1 + {{\tan }^2}t} \right)dt\)
\[ \Rightarrow I = \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\left( {1 + {{\tan }^2}t} \right)dt}}{{1 + {{\tan }^2}t}}} = \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {dt} = \left. t \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} = \dfrac{\pi }{3} - \dfrac{\pi }{6} = \dfrac{\pi }{6}\]
\(e)\int\limits_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx} \)
Đặt \(x = \sin t \Rightarrow dx = \cos tdt\)
Đổi cận \(x = - \dfrac{1}{2} \Rightarrow t = - \dfrac{\pi }{6},x = \dfrac{1}{2} \Rightarrow t = \dfrac{\pi }{6}\)
\[ \Rightarrow I = \int\limits_{ - \dfrac{\pi }{6}}^{\dfrac{\pi }{6}} {\dfrac{{\cos tdt}}{{\sqrt {1 - {{\sin }^2}t} }}} = \int\limits_{ - \dfrac{\pi }{6}}^{\dfrac{\pi }{6}} {\dfrac{{\cos tdt}}{{\cos t}}} = \int\limits_{ - \dfrac{\pi }{6}}^{\dfrac{\pi }{6}} {dt} = \left. t \right|_{ - \dfrac{\pi }{6}}^{\dfrac{\pi }{6}} = \dfrac{\pi }{6} + \dfrac{\pi }{6} = \dfrac{\pi }{3}\]
