Question

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

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

Answer 1

Answer: (B)

$I_n=\int {\cot }^{n}xdx=\int {\cot }^{n-2}\cdot x\cdot \left({\mathrm{cosec}}^{2}x-1\right)dx$