∫2cosec⁡2x1+cot8⁡xdx=−18log⁡(f(x))+c then f(x)=

I=∫2cosec⁡2x1+cot8⁡xdxI=∫1sin⁡xcos⁡x1+cot8⁡xdxI=∫cot7⁡xsin2⁡xcot8⁡x1+cot8⁡xdx⇒−18∫−8cot7⁡xcosec2⁡xcot8⁡x1+cot8⁡xdx Let cot8⁡x=t−8cot7⁡xcos⁡ec2xdx=dtI=−18∫1t(1+t)dt=−18∫1t−1t+1dt=-18log⁡t+1t+C=-18logcos8x+sin8xcos8x+c

∫2cosec⁡2x1+cot8⁡xdx=−18log⁡(f(x))+c then f(x)=

I=∫2cosec⁡2x1+cot8⁡xdxI=∫1sin⁡xcos⁡x1+cot8⁡xdxI=∫cot7⁡xsin2⁡xcot8⁡x1+cot8⁡xdx⇒−18∫−8cot7⁡xcosec2⁡xcot8⁡x1+cot8⁡xdx Let cot8⁡x=t−8cot7⁡xcos⁡ec2xdx=dtI=−18∫1t(1+t)dt=−18∫1t−1t+1dt=-18log⁡t+1t+C=-18logcos8x+sin8xcos8x+c

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$\int \frac{2 cosec 2 x}{(1 + \cot^{8} x)} d x = - \frac{1}{8} \log (f (x)) + c then f (x) =$

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$\frac{\cos^{8} x + \sin^{8} x}{\cos^{8} x}$

$\frac{\cos^{8} x + \sin^{8} x}{\sin^{8} x}$

$\frac{\cos^{8} x - \sin^{8} x}{\sin^{8} x}$

$\frac{\cos^{8} x - \sin^{8} x}{\cos^{8} x}$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int \frac{2 cosec 2 x}{(1 + \cot^{8} x)} d x \\ I = \int \frac{1}{\sin x \cos x (1 + \cot^{8} x)} d x \\ I = \int \frac{\cot^{7} x}{\sin^{2} x \cot^{8} x (1 + \cot^{8} x)} d x \\ \Rightarrow - \frac{1}{8} [\int \frac{- 8 \cot^{7} x {cosec}^{2} x}{\cot^{8} x (1 + \cot^{8} x)} d x] \\ Let \cot^{8} x = t \\ - 8 \cot^{7} x \cos e c^{2} x d x = d t \\ I = \frac{- 1}{8} \int \frac{1}{t (1 + t)} d t \\ = - \frac{1}{8} \int \frac{1}{t} - \frac{1}{t + 1} d t \\ = - \frac{1}{8} \log (\frac{t + 1}{t}) + C \end{array}$