MCQ
$f\left( x \right) = \int {\frac{{dx}}{{{{\sin }^6}\,x}}} $ બહુપદીએ . . . ઘાતાંક . . .માં છે .
- ✓$5$ , $cot\, x$
- B$5$ , $tan\, x$
- C$3$ , $tan\, x$
- D$3$ , $cot\, x$
$f(x)=\int \csc ^{6} x d x$
From reduction formula, we have
$I_{n}=\int \csc ^{n} x d x$
$=-\frac{\csc ^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} 1_{n-2}$
$\therefore f(x)=-\frac{\csc ^{4} x \cot x}{5}+\frac{4}{5}\left[\begin{array}{cc}{-\csc ^{2} x \cot x} & {+\frac{2}{3} I_{2}} \\ {3} & {3}\end{array}\right]$
$=\frac{\csc ^{4} x \cot x}{5}-\frac{4}{15} \csc ^{2} x \cdot \cot x+\frac{8}{15}[-\cot x]$
$=\frac{-\left(1+\cot ^{2} x\right)^{2} \cdot \cot x}{5}-\frac{4}{15}\left(1+\cot ^{2} x\right) \cot x$
$-\frac{8}{15}(-\cot x)\left(\because \csc ^{2} x=1+\cot ^{2} x\right)$
$\begin{aligned}=& \frac{-1}{5}\left[1+\cot ^{4} x+2 \cot ^{2} x\right] \cot x-\frac{4}{15}\left[\cot x+\cot ^{3} x\right] \\ &-\frac{8}{15} \cot x \\=& \frac{-1}{5}\left[\cot x+\cot ^{5} x+2 \cot ^{3} x\right] \\=& \frac{-15}{15} \cot x-\frac{\cot ^{5} x}{5}-\frac{10}{15} \cot ^{3} x \\=& \frac{-\cot ^{5} x}{5}-\frac{2}{3} \cot ^{3} x-\cot x \end{aligned}$
It is a polynomial of degree $5$ in cot $x$.
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