Cho \(F(x)\) là một nguyên hàm của hàm số \(y = \frac{1}{{1 + \sin 2x}}\) với \(x \in R\backslash \left\{ { - \frac{\pi }{4} + k

* Hướng dẫn giải

Với \(x\) thuộc vào mỗi khoảng \(\left( { - \frac{\pi }{4} + k\pi ; - \frac{\pi }{4} + k\pi } \right),{\rm{ }}k \in Z\) ta có

\(F\left( x \right) = \int {\frac{{{\rm{d}}x}}{{1 + \sin 2x}} = \int {\frac{{{\rm{d}}x}}{{{{\left( {\sin x + \cos x} \right)}^2}}} = \int {\frac{{{\rm{d}}x}}{{2{{\cos }^2}\left( {x + \frac{\pi }{4}} \right)}} = \frac{1}{2}tan\left( {x + \frac{\pi }{4}} \right)} } }  + C.\)

 \(0; - \frac{\pi }{{12}} \in \left( { - \frac{\pi }{4};\frac{\pi }{4}} \right)\) nên \(F\left( 0 \right) - F\left( { - \frac{\pi }{{12}}} \right) = \frac{1}{2}\tan \left( {x - \frac{\pi }{4}} \right)\left| {\begin{array}{*{20}{c}}
0\\
{ - \frac{\pi }{{12}}}
\end{array}} \right. =  - \frac{1}{2} + \frac{{\sqrt 3 }}{2}\mathop  \to \limits^{F(0) = 1} F\left( { - \frac{\pi }{{12}}} \right) = \frac{3}{2} - \frac{{\sqrt 3 }}{2}.\)

\(\pi ;\frac{{11\pi }}{{12}} \in \left( {\frac{\pi }{4};\frac{{5\pi }}{4}} \right)\) nên \(F\left( \pi  \right) - F\left( {\frac{{11\pi }}{{12}}} \right) = \frac{1}{2}\tan \left( {x - \frac{\pi }{4}} \right)\left| {\begin{array}{*{20}{c}}
\pi \\
{\frac{{11\pi }}{{12}}}
\end{array}} \right. =  - \frac{1}{2} + \frac{{\sqrt 3 }}{2}\mathop  \to \limits^{F\left( \pi  \right) = 0} F\left( {\frac{{11\pi }}{{12}}} \right) = \frac{1}{2} - \frac{{\sqrt 3 }}{2}.\)

Vậy \(P = F\left( { - \frac{\pi }{{12}}} \right) - F\left( {\frac{{11\pi }}{{12}}} \right) = 1.\)