[tex]=\int x(-\cot x)'\ dx=-x\cot x +\int \cot x \ dx[/tex]
Luam a doua integrala si o rezolvam prin schimbare de variabila
[tex]t=\sin x\\ dt=\cos x \ dx[/tex]
[tex]\int \cot x \ dx=\int\dfrac{\cos x}{\sin x}dx=\int\dfrac{dt}{t}=\ln t=\ln(\sin x).[/tex]
Raspunsul final e:
[tex]-x\cot x+\ln(\sin x)+C[/tex]
