I want to calculate $$I_n = \int \frac{d\theta}{\sin^n(c\theta)\cdot \cos(c\theta)}. $$ The answer is $$-\frac{1}{c(n-1)\sin^{n-1}(c\theta)}+ \int\frac{d\theta}{\sin^{n-2}(c\theta)\cdot\cos(c\theta)}.$$ I started on this way: $$ I_n = \frac{1}{c} \int \frac{c\cdot \cos (c\theta) \, d\theta}{\sin^n(c\theta) \cdot \cos(c\theta)} = \frac{1}{c} \int \frac{dt}{t^n \cdot (1-t^2)},$$ being $t=\sin(c\theta)$.
Am I on the correct way? Any suggestions and hints? Thanks a lot!
Let $t=c\theta$, $dt =c d\theta$ to get $I_n=\frac{1}{c}\int\csc^{n}t\sec t dt$.
Now let $u=\sec t\csc^{n-2}t, dv=\csc^{2}t dt$, so that
$du=[\sec t\tan t\csc^{n-2}t+\sec t[(n-2)\csc^{n-3}t\;(-\csc t\cot t)]dt, \;\;v=-\cot t$,
to get $cI_n=-\sec t\csc^{n-2}t\cot t+\int\sec t\csc^{n-2}t\;dt-(n-2)\int\sec t \csc^{n-2}t\cot^{2}t\; dt$
$=-\csc^{n-1}t+\int\sec t\csc^{n-2}t\;dt-(n-2)\int\sec t\csc^{n-2}t(\csc^{2}t-1)\;dt$
$=-\csc^{n-1}t+(n-1)\int\sec t\csc^{n-2}t\;dt-(n-2)cI_n$.
Then $(n-1)cI_n=-\csc^{n-1}t+(n-1)\int\sec t\csc^{n-2}t \;dt$,
so $\displaystyle I_n=-\frac{\csc^{n-1}(c\theta)}{c(n-1)}+\int\sec (c\theta)\csc^{n-2}(c\theta)\; d\theta$.