Exercice 55

\(\begin{align*} \int \frac{1-tanx}{1-tanx}dx \end{align*}\)

Connaissances:

  • trigonométrie
  • changement de variable
  • primitive de \(u'/u\)

\( \begin{align*} I & = \int \frac{1-tanx}{1+tanx}dx \\ 
& = \int \frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}dx \\ 
& = \int \frac{\frac{cosx - sinx}{cosx}}{\frac{cosx+sinx}{cosx}}dx \\ 
& = \int \frac{cosx - sinx}{cosx+sinx}dx \\
& u=cosx+sinx \Rightarrow du=-sinx+cosx \\
I & = \int \frac{du}{u} \\
& = ln \lvert u \rvert + C \\
& = ln \lvert cosx+sinx \rvert + C \\ \end{align*}\)

\[\boxed {\begin{align*} I = ln \lvert cosx+sinx \rvert + C(\in \mathbb R) \end{align*}}\]