Tích phân \(I = \int\limits_{ - \frac{\pi }{3}}^{\frac{\pi }{3}} {\frac{{\sin x}}{{{{\left( {\cos x + \sqrt 3 \sin x} \right)}^2}}}dx} \) có gái trị là:

* Hướng dẫn giải

Ta có:

\(\begin{array}{l} I = \int\limits_{ - \frac{\pi }{3}}^{\frac{\pi }{3}} {\frac{{\sin x}}{{{{\left( {\cos x + \sqrt 3 \sin x} \right)}^2}}}dx} \\ = \int\limits_{ - \frac{\pi }{3}}^{\frac{\pi }{3}} {\frac{{\sin x}}{{4{{\left( {\frac{1}{2}\cos x + \frac{{\sqrt 3 }}{2}\sin x} \right)}^2}}}dx} I\\ = \int\limits_{ - \frac{\pi }{3}}^{\frac{\pi }{3}} {\frac{{\sin x}}{{4{{\left[ {\sin \left( {x + \frac{\pi }{6}} \right)} \right]}^2}}}dx} \end{array}\).

Đặt \(u = x + \frac{\pi }{6} \Rightarrow x = u - \frac{\pi }{6} \Rightarrow dx = du\).

Đổi cận \(\left\{ \begin{array}{l} x = - \frac{\pi }{3} \Rightarrow u = - \frac{\pi }{6}\\ x = \frac{\pi }{3} \Rightarrow u = \frac{\pi }{2} \end{array} \right.\)

\(\begin{array}{l} I = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\sin \left( {u - \frac{\pi }{6}} \right)}}{{4{{\sin }^2}u}}du} \\ = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\sin u.\cos \frac{\pi }{6} - \sin \frac{\pi }{6}\cos u}}{{4{{\sin }^2}u}}du} \\ = \frac{1}{8}\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\sqrt 3 .\sin u - \cos u}}{{{{\sin }^2}u}}du} \\ = \frac{1}{8}\left( {\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\sqrt 3 \sin u}}{{1 - {{\cos }^2}u}}du - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\cos u}}{{{{\sin }^2}u}}du} } } \right) \end{array}\)

Xét \({I_1} = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\sqrt 3 \sin u}}{{1 - {{\cos }^2}u}}du} \).

Đặt \(t = \cos u,u \in \left[ {0;\pi } \right] \Rightarrow dt = - \sin udu\).

Đổi cận \(\left\{ \begin{array}{l} u = - \frac{\pi }{6} \Rightarrow t = \frac{{\sqrt 3 }}{2}\\ u = \frac{\pi }{2} \Rightarrow t = 0 \end{array} \right.\).

\(\begin{array}{l} \Rightarrow {I_1} = \int\limits_{\frac{{\sqrt 3 }}{2}}^0 {\frac{{\sqrt 3 dt}}{{1 - {t^2}}}} \\ = \frac{{\sqrt 3 }}{2}\int\limits_{\frac{{\sqrt 3 }}{2}}^0 {\left( {\frac{1}{{1 - t}} + \frac{1}{{1 + t}}} \right)} dt\\ = \frac{{\sqrt 3 }}{2}\left. {\left( {ln\left| {\frac{{t + 1}}{{t - 1}}} \right|} \right)} \right|_{\frac{{\sqrt 3 }}{2}}^0\\ = - \frac{{\sqrt 3 }}{2}\ln \left( {\frac{{\sqrt 3 + 2}}{{ - \sqrt 3 + 2}}} \right) \end{array}\)

Xét \({I_2} = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\cos u}}{{{{\sin }^2}u}}du} \).

Đặt \(t = \sin u,u \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right] \Rightarrow dt = \cos udu\).

Đổi cận \(\left\{ \begin{array}{l} u = - \frac{\pi }{6} \Rightarrow t = - \frac{1}{2}\\ u = \frac{\pi }{2} \Rightarrow t = 1 \end{array} \right.\).

\({I_2} = \int\limits_{ - \frac{1}{2}}^1 {\frac{1}{{{t^2}}}du} = \left. {\left( { - \frac{1}{t}} \right)} \right|_{ - \frac{1}{2}}^1 = - 3\).

\(\Rightarrow I = \frac{1}{8}\left( {{I_1} - {I_2}} \right) = - \frac{{\sqrt 3 }}{{16}}\ln \left( {\frac{{\sqrt 3 + 2}}{{ - \sqrt 3 + 2}}} \right) + \frac{3}{8}\).