Question

Obtain the reduction formula for $I_n=\int cose{c}^{n}xdx$, $n$ being a positive integer, $n\ge 2$ and hence deduce the value of $\int cose{c}^{5}xdx$.

Answer 1

$I_n=\int {\csc }^{n}x=\int {\csc }^{n-2}x{\csc }^{2}x$

By integration by parts , we get $I_n={\csc }^{n-2}x(-cotx)-\int (-cotx)(n-2){\csc }^{n-3}x(-cscx)cotxdx=-{\csc }^{n-2}xcotx-(n-2)\int {\csc }^{n-2}x{\cot }^{2}x$

$I_n=-{\csc }^{n-2}xcotx-(n-2)\int {\csc }^{n-2}x({\csc }^{2}x-1)=\frac{{\csc }^{n-2}xcotx}{n-1}+\frac{n-2}{n-1}\int {\csc }^{n-2}x$

Therefore the value of $\int {\csc }^{5}x=\frac{{\csc }^{3}xcotx}{4}+\frac{3}{4}\int {\csc }^{3}x$

$=\frac{{\csc }^{3}xcotx}{4}+\frac{3}{4}(\frac{cscxcotx}{2}+\frac{1}{2}\int cscx)$

$=\frac{{\csc }^{3}xcotx}{4}+\frac{3cscxcotx}{4}+\frac{-3\ln (cscx+cotx)}{8}$