A) 5 in cot x
B) 5 in tan x
C) 3 in tan x
D) 3 in cot x
Correct Answer: A
Solution :
Let\[f(x)=\int_{{}}^{{}}{\frac{dx}{{{\sin }^{6}}x}}\] \[f(x)=\int_{{}}^{{}}{\cos e{{c}^{6}}xdx}\] From reduction formula, we have \[{{I}_{n}}=\int_{{}}^{{}}{\cos e{{c}^{n}}xdx}\] \[=-\frac{\cos e{{c}^{n-2}}x\cot x}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}\] \[\therefore \]\[f(x)=-\frac{\cos e{{c}^{4}}x\cot x}{5}+\frac{4}{5}\left[ \frac{-\cos e{{c}^{2}}x\cot x}{3}+\frac{2}{3}{{I}_{2}} \right]\]\[=-\frac{\cos e{{c}^{4}}x\cot x}{5}-\frac{4}{5}\cos e{{c}^{2}}x.\cot x+\frac{8}{15}[-\cot x]\]\[=\frac{-{{(1+{{\cot }^{2}}x)}^{2}}.\cot x}{5}-\frac{4}{15}(1+{{\cot }^{2}}x)\cot x\] \[-\frac{8}{15}(-\cot x)(\because \cos e{{c}^{2}}x=1+{{\cot }^{2}}x)\] \[=\frac{-1}{5}\left[ 1+{{\cot }^{4}}x+2{{\cot }^{2}}x \right]\cot x-\frac{4}{15}\left[ \cot x+{{\cot }^{3}}x \right]\]\[-\frac{8}{15}\cot x\] \[=\frac{-1}{5}[\cot x+{{\cot }^{5}}x+2{{\cot }^{3}}x]\] \[=\frac{-4}{5}\cot x-\frac{4}{15}{{\cot }^{3}}x-\frac{8}{15}\cot x\] \[=\frac{-15}{15}\cot x-\frac{{{\cot }^{5}}x}{5}-\frac{10}{15}{{\cot }^{3}}x\] \[=\frac{-{{\cot }^{5}}x}{5}-\frac{2}{3}{{\cot }^{3}}x-\cot x\] It is a polynomial of degree 5 in cot x.
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