Chủ đề này có 1 trả lời

[Mr0]

$\int \frac{dx}{\sqrt{2}+sinx-cosx}$

[Crystal]

$\int \frac{dx}{\sqrt{2}+sinx-cosx}$

Ta có: \[I = \int {\frac{{dx}}{{\sqrt 2 + sinx - cosx}} = \int {\frac{{dx}}{{\sqrt 2 \left( {1 - \cos \left( {x + \frac{\pi }{4}} \right)} \right)}}} } = \int {\dfrac{{dx}}{{2\sqrt 2 {{\sin }^2}\left( {\frac{x}{2} + \dfrac{\pi }{8}} \right)}}} \]
\[ = \frac{1}{{2\sqrt 2 }}\int {\dfrac{{dx}}{{{{\sin }^2}\left( {\dfrac{x}{2} + \dfrac{\pi }{8}} \right)}}} = \dfrac{1}{{\sqrt 2 }}\int {\dfrac{{d\left( {\dfrac{x}{2} + \dfrac{\pi }{8}} \right)}}{{{{\sin }^2}\left( {\dfrac{x}{2} + \dfrac{\pi }{8}} \right)}}} = - \dfrac{1}{{\sqrt 2 }}\cot g\left( {\dfrac{x}{2} + \dfrac{\pi }{8}} \right) + C\]