Question
$\int \frac{1}{x \sin ^2(\log x)} d x$

Answer

$\text { Let } I =\int \frac{1}{x \cdot \sin ^2(\log x)} d x$
$\text { Put } \log x = t$
$\therefore \frac{1}{x} d x= dt$
$\therefore I =\int \frac{1}{\sin ^2 t } dt$
$=\int \operatorname{cosec}{ }^2 t \cdot dt$
$=-\cot t + c$
$\therefore I =-\cot (\log x )+ c $

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