[tex]\displaystyle\dfrac{1}{\sec x(1-\sin x)}=\dfrac{\cos x}{1-\sin x}\\\\\\\int \dfrac{dx}{\sec x(1-\sin x)}=\int\dfrac{\cos x\, dx}{1-\sin x}\\\\\\u=1-\sin x\\du=-\cos x\, dx\\\\\int \dfrac{\cos x\, dx}{1-\sin x}=\int \dfrac{-du }{u}=-\int \dfrac{du}{u}=-\ln |u|+C=-\ln|1-\sin x|+C[/tex]
