Đáp án:
$\begin{array}{l}
Do:\left( {\cos x + \sin x} \right)' = - \sin x + \cos x\\
\Rightarrow \int {\frac{{{\mathop{\rm cosx}\nolimits} - sinx}}{{\cos x + \sin x}}dx} \\
= \int {\frac{1}{{\cos x + \sin x}}.\left( {{\mathop{\rm cosx}\nolimits} - sinx} \right)dx} \\
= \int {\frac{1}{{\cos x + \sin x}}.d\left( {\cos x + \sin x} \right)} \\
= \ln \left| {\sin x + \cos x} \right| + C
\end{array}$