Câu hỏi 35

Ta có :

\(\begin{array}{l}\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{1}{{1 + \tan x}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{1}{{1 + \dfrac{{\sin x}}{{\cos x}}}}dx} \\ = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{1}{{\dfrac{{\cos x + \sin x}}{{\cos x}}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\cos x}}{{\sin x + \cos x}}dx} \\ = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\sin x + \cos x + \cos x - \sin x}}{{2\left( {\sin x + \cos x} \right)}}dx} \\ = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{4}} {\left( {1 + \dfrac{{\cos x - \sin x}}{{\sin x + \cos x}}} \right)dx} \\ = \dfrac{1}{2}\left[ {x + \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{d\left( {\sin x + \cos x} \right)}}{{\sin x + \cos x}}} } \right]\\ = \dfrac{1}{2}\left. {\left( {x + \ln \left| {\sin x + \cos x} \right|} \right)} \right|_0^{\dfrac{\pi }{4}}\\ = \dfrac{1}{2}\left( {\dfrac{\pi }{4} + \ln \sqrt 2 } \right) = \dfrac{\pi }{8} + \dfrac{1}{2}\ln \sqrt 2 \\ = \dfrac{\pi }{8} + \dfrac{1}{4}\ln 2\\ \Rightarrow a = \dfrac{1}{8},b = \dfrac{1}{4} \Rightarrow \dfrac{a}{b} = \dfrac{1}{2}\end{array}\)

Chọn A.