Lời giải chi tiết:

\(\begin{array}{l}
f\left( x \right) = \dfrac{1}{{1 + \sin x}} = \dfrac{1}{{{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}\\
\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{{{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}^2}}} = \dfrac{1}{{{{\left( {\sqrt 2 \left( {\sin \dfrac{x}{2} + \dfrac{\pi }{4}} \right)} \right)}^2}}} = \dfrac{1}{{2{{\sin }^2}\left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)}}\\
\Rightarrow F\left( x \right) = \dfrac{1}{2}\int {\dfrac{1}{{{{\sin }^2}\left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)}}dx} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 1}}{2}.2\cot \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right) + C = - \cot \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right) + C\\
Khi\,\,\,{\mkern 1mu} C = 1 \Rightarrow F\left( x \right) = 1 - \cot \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)\end{array}\)

Chọn A.