Question

If $I_n=\int {\cot }^{n}xdx$, then $I_0+I_1+2(I_2+I_3+....+I_8)+I_9+I_{10}$ equals to: (where $u=\cot x)$.

• A. $u+\frac{u^2}{2}+....+\frac{u^9}{9}$
• B. $-(u+\frac{u^2}{2}+....+\frac{u^9}{9})$
• C. $-(u+\frac{u^2}{2!}+....+\frac{u^9}{9!})$
• D. $\frac{u}{2}+\frac{2u^2}{3}+.....+\frac{9u^9}{10}$

Answer 1

The correct option is B $-(u+\frac{u^2}{2}+....+\frac{u^9}{9})$

$I_n=\int {\cot }^{n}xdx=\int {\cot }^{n-2}x{\cot }^{2}xdx=\int {\cot }^{n-2}x{\text{cosec}}^{2}xdx-\int {\cot }^{n-2}xdx$

$I_n+I_{n-2}=-\frac{{\cot }^{n-1}x}{n-1}$

$I_0+I_1+2(I_3+I_4+\cdots +I_8)+I_9+I_{10}=(I_2+I_0)+(I_3+I_1)+\cdots +(I_9+I_7)+(I_{10}+I_8)$

$=-(u+\frac{u^2}{2}+\cdots +\frac{u^9}{9})$