Question

$f\left(x\right) = \int \frac{dx}{sin^{6} x}$ 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 $f\left(x\right) = \int \frac{dx}{sin^{6} x}$
$f\left(x\right) = \int cosec^{6}x\, dx$
From reduction formula, we have
$I_{n} = \int cosec^{n}x\, dx$
$=- \frac{cosec^{n-2}x\,cot\,x}{n-1}+\frac{n-2}{n-1}I_{n-2}$
$\therefore f\left(x\right) = \frac{cosec^{4}x\,cot\,x}{5}+\frac{4}{5}\left[\frac{-cosec^{2}x\,cot\,x}{3}+\frac{2}{3}I_{2}\right]$
$= \frac{cosec^{4}x\,cot\,x}{5}-\frac{4}{5}cosec^{2}x\,cot\,x+\frac{8}{15}\left[-cot\,x\right]$
$= \frac{-\left(1 + cot^{2} x\right)2. cot x}{5} -\frac{4}{15}\left(1 + cot^{2} x\right) cot \,x$
$-\frac{8}{15}\left(-cot\,x\right) \left(\because cosec^{2} x = 1 + cot^{2} x\right)$
$= \frac{-1}{5}\left[1 + cot^{4}x + 2\,cot^{2}x\right]cot\, x -\frac{4}{15}\left[cot \,x + cot^{3} x\right]$
$-\frac{8}{15}cot\, x$
$= \frac{-1}{5}\left[cot\,x + cot^{5}x + 2\,cot^{2}x\right]$
$\frac{-4}{15}cot\, x -\frac{4}{15}cot^{3}\, x-\frac{8}{15}cot\, x$
$= \frac{-15}{15}cot\, x-\frac{cot^{5}x}{5}-\frac{10}{15}cot^{3}\, x$
$= \frac{-cot^{5}x}{5}-\frac{2}{3}cot^{3}\, x-cot\, x$
It is a polynomial of degree 5 in cot x.