Question
$\int[\operatorname{cosec}(\log x)][1-\cot (\log x)] d x$

Answer

$\text { Let } I =\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$
$=\int\left(\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cdot \cos ^2 x}\right) d x \ldots \ldots\left[\because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta\right]$
$=\int\left(\frac{\cos ^2 x}{\sin ^2 x \cdot \cos ^2 x}-\frac{\sin ^2 x}{\sin ^2 x \cdot \cos ^2 x}\right) d x$
$=\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x$
$=\int \operatorname{cosec}^2 x d x-\int \sec ^2 x d x$
$\therefore I =-\cot x -\tan x + c $

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