$\int \frac{cot (log x)}{x} d x$
is equal to

log (sin x) + C

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log (log x) + C

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log [sin (log x)] + C

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sin [log (log x)] + C

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Solution

The correct option is A $log (sin x) + C$
$= \int \frac{cot (log x)}{x}$
$\Rightarrow log x = t$
$\int cot t = log (sin x) + c$

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Similar questions

\int [log (log x) + \frac{1}{{(log x)}^{2}}] d x = x [f (x) - g (x)] + c

The derivative of log [log(logx)]w.r.t x is:

\int \frac{1}{\sin (x - a) \sin (x - b)} d x i s e q u a l t o (a) \sin (b - a) \log |\frac{\sin (x - a)}{\sin (x - b)}| + C (b) \cos e c (b - a) \log |\frac{\sin (x - a)}{\sin (x - b)}| + C (c) \cos e c (b - a) \log |\frac{\sin (x - b)}{\sin (x - a)}| + C (d) \sin (b - a) \log |\frac{\sin (x - a)}{\sin (x - b)}| + C

(e, \infty) \to R

f (x) = log [log (log x)]

f

y = log [log (log x)]

\frac{d y}{d x}

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