Given:
[tex]\int (\frac{\sin 3t}{\cos 3t-1})\differentialDt t[/tex]Sol:.
[tex]\begin{gathered} \int \frac{\sin3t}{\cos3t-1}\differentialDt t \\ u=\cos 3t-1 \\ \frac{du}{dt}=-3\sin 3t \\ \differentialDt t=\frac{-1}{3\sin 3t}du \\ =-\frac{1}{3}\int \frac{1}{u}du \end{gathered}[/tex][tex]\begin{gathered} \int \frac{1}{u}du \\ =\ln (u) \end{gathered}[/tex][tex]\begin{gathered} =-\frac{1}{3}\int \frac{1}{u}du \\ =-\frac{\ln u}{3} \\ u=\cos 3x-1 \\ =-\frac{\ln (1-\cos 3x)}{3}+c \end{gathered}[/tex]