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

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

$=\int {\cot }^{n-2}x({\csc }^{2}x-1)dx=\int {\cot }^{n-2}x{\cos }^{2}xdx-I_{n-2}$

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

$\therefore I_0+I_1+2(I_2+...I_8)+I_9+I_{10}$

$=(I_2+I_0)+(I_3+I_1)+(I_4+I_2)+(I_5+I_3)+(I_6+I_4)+(I_7+I_5)$

$=-(\frac{\cos x}{1}+\frac{{\cos }^{2}x}{2}+...+\frac{{\cos }^{9}x}{9})+C$