Questions mathématiques diverses

Question 6:
Calculer : $I=\int_0^{\frac{\pi}{6}}\dfrac{\sin x}{\sin x-\cos x}dx$

Réponse 6:
$$\begin{align*} &\text{Pour tout réel x, on peut écrire}\\ &\sin x=-\sin x+\cos x+\sin x+\sin x-\cos x\\ &I=-\int_{0}^{\frac{\pi}{6}}{\dfrac{\sin x}{\sin x-\cos x}dx}+\int_{0}^{\frac{\pi}{6}}{\dfrac{\cos x+\sin x}{\sin x-\cos x}dx}+\int_{0}^{\frac{\pi}{6}}{\dfrac{\sin x-\cos x}{\sin x-\cos x}dx}\\ &\iff I=-I+{\left[{\ln|\sin x-\cos x|}\right]}_{0}^{\frac{\pi}{6}}+\left[{x}\right]_{0}^{\frac{\pi}{6}}\\ &\iff I=-I+\ln\left({\cos \frac{\pi}{6}-\sin \frac{\pi}{6}}\right)+\frac{\pi}{6}\\ &\iff I=-I+\ln\left({\dfrac{\sqrt{3}-1}{2}}\right)+\dfrac{\pi}{6}\\ &\iff 2I=\ln\left({\dfrac{\sqrt{3}-1}{2}}\right)+\dfrac{\pi}{6}\\ &\iff I=\dfrac{1}{2}\ln\left({\dfrac{\sqrt{3}-1}{2}}\right)+\dfrac{\pi}{12} \end{align*}$$