$$\frac{I_n}{a}=\int sin^n(cx)dx=\int sin^{n-1}(cx).sin(cx)dx=$$, Eğer $$u=sin^{n-1}(cx)\Rightarrow du=c.(n-1)sin^{n-2}(cx).cos(cx)dx$$
$$sin(cx)dx=dv\rightarrow v=-\frac{1}{c}cos(cx)$$,
$$=-\frac{1}{c}.cos(cx)sin^{n-1}(cx)-\int (n-1)sin^{n-2}(cx).cos^2(cx)dx $$
$$=-\frac{1}{c}.cos(cx)sin^{n-1}(cx)+(1-n)\int sin^{n-2}(cx).(1-sin^2(cx))dx $$
$$=-\frac{1}{c}.cos(cx)sin^{n-1}(cx)+(1-n)\int sin^{n-2}(cx)dx- (1-n)\int sin^n(cx)dx $$
$$=-\frac{1}{c}.cos(cx )sin^{n-1}(cx)+(1-n)\frac{I_{n-2}}{a}+ (n-1)\frac{I_n}{a} $$
$$I_n=-\frac{a}{c(2-n)}.cos(cx )sin^{n-1}(cx)+\frac{(1-n)}{(2-n)}.I_{n-2} $$