Question

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

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

Answer 1

Answer: (B)

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