int frac{cot x}{log(sin x)} , dx = | vedclass

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Question

(A) Let $I = \int \frac{\cot x}{\log(\sin x)} \, dx$. Substitute $t = \log(\sin x)$. Differentiating both sides with respect to $x$,we get: $dt = \fra

Vedclass · Accepted Answer

$\log(\log(\sin x)) + c$

Vedclass · Answer

(A) Let $I = \int \frac{\cot x}{\log(\sin x)} \, dx$. Substitute $t = \log(\sin x)$. Differentiating both sides with respect to $x$,we get: $dt = \frac{1}{\sin x} \cdot \cos x \, dx = \cot x \, dx$. Substituting these into the integral,we get: $I = \int \frac{1}{t} \, dt$. Integrating with respect to $t$,we get: $I = \log|t| + c$. Substituting back the value of $t$,we get: $I = \log|\log(\sin x)| + c$.

$\int \frac{\cot x}{\log(\sin x)} \, dx = $

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