Question

$f\left(x\right)=\int \frac{dx}{si{n}^{6}x}$ is a polynomial of degree

• A. 5 in cot x
• B. 5 in tan x
• C. 3 in tan x
• D. 3 in cot x

Answer 1

Answer: (A)

Let $f\left(x\right)=\int \frac{dx}{si{n}^{6}x}$
$f\left(x\right)=\int cose{c}^{6}xdx$
From reduction formula, we have
$I_n=\int cose{c}^{n}xdx$
$=-\frac{cose{c}^{n-2}xcotx}{n-1}+\frac{n-2}{n-1}{I}_{n-2}$
$\therefore f\left(x\right)=\frac{cose{c}^{4}xcotx}{5}+\frac{4}{5}\left[\frac{-cose{c}^{2}xcotx}{3}+\frac{2}{3}{I}_2\right]$
$=\frac{cose{c}^{4}xcotx}{5}-\frac{4}{5}cose{c}^{2}xcotx+\frac{8}{15}\left[-cotx\right]$
$=\frac{-\left(1+co{t}^{2}x\right)2.cotx}{5}-\frac{4}{15}\left(1+co{t}^{2}x\right)cotx$
$-\frac{8}{15}\left(-cotx\right)\left(\because cose{c}^{2}x=1+co{t}^{2}x\right)$
$=\frac{-1}{5}\left[1+co{t}^{4}x+2co{t}^{2}x\right]cotx-\frac{4}{15}\left[cotx+co{t}^{3}x\right]$
$-\frac{8}{15}cotx$
$=\frac{-1}{5}\left[cotx+co{t}^{5}x+2co{t}^{2}x\right]$
$\frac{-4}{15}cotx-\frac{4}{15}co{t}^{3}x-\frac{8}{15}cotx$
$=\frac{-15}{15}cotx-\frac{co{t}^{5}x}{5}-\frac{10}{15}co{t}^{3}x$
$=\frac{-co{t}^{5}x}{5}-\frac{2}{3}co{t}^{3}x-cotx$
It is a polynomial of degree 5 in cot x.