Question

lf $I_n=\int {\cot }^{n}xdx$ then $I_4+I_6=$

• A. $-\frac{{\cot }^{5}x}{5}$
• B. ${\cot }^{5}x$
• C. $\frac{1}{5}{\cot }^{5}x$
• D. $-\frac{1}{5}{\cot }^{5}x-\frac{1}{3}{\cot }^{3}x$

Answer 1

Answer: (A)

$-\frac{{\cot }^{5}x}{5}$

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

$=\int {\cot }^{n-2}x[cose{c}^{2}x-1]dx$

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

$\Rightarrow I_n+I_{n-2}=-\int {\cot }^{n-2}xd\left(\cot x\right)$

$=-\frac{{\cot }^{(n-2+1)}x}{n-1}$

$=\frac{-{\cot }^{(n-1)}x}{n-1}$

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

$n=6=I_6+I_4=\frac{-{\cot }^{5}x}{5}$