Question

If $I_n=\int {\cot }^{n}xdx$ and $I_0+I_1+2(I_2+...+I_8)+I_9+I_{10}=A(u+\frac{u^2}{2}+...+\frac{u^9}{9})+$ constant where $u=\cot x$ then,

• A. $A$ is constant
• B. $A=-1$
• C. $A=1$
• D. $A$ is dependent on $x$

Answer 1

Answer: (A), (B)

The correct options are
A $A$ is constant
B $A=-1$

$I_n=\int {\cot }^{n}xdx=\int {\cot }^{n-2}x{\cot }^{2}xdx=\int {\cot }^{n-2}x({\csc }^{2}x-1)dx=\int {\cot }^{n-2}x{\csc }^{2}xdx-I_{n-2}$
$\Rightarrow I_n+I_{n-2}=-\frac{{\cot }^{n-1}x}{n-1}+c$
$I_0+I_1+2(I_2+...I_8)+I_9+I_{10}=(I_2+I_0)+(I_3+I_1)+(I_4+I_2)+(I_5+I_3)+(I_6+I_4)+(I_7+I_5)+(I_8+I_6)+(I_9+I_7)+(I_{10}+I_8)$
$=-(\frac{\cot x}{1}+\frac{{\cot }^{2}x}{2}+...+\frac{{\cot }^{9}x}{9})+c$
Hence $A=-1$