int cot x cdot log [log (sin x)] d x= | vedclass

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Question

(C) Let $I = \int \cot x \cdot \log [\log (\sin x)] d x$. Substitute $t = \log (\sin x)$. Then $dt = \frac{1}{\sin x} \cdot \cos x \, dx = \cot x \, d

Vedclass · Accepted Answer

$\log (\sin x)[\log (\log (\sin x))-1]+c$

Vedclass · Answer

(C) Let $I = \int \cot x \cdot \log [\log (\sin x)] d x$. Substitute $t = \log (\sin x)$. Then $dt = \frac{1}{\sin x} \cdot \cos x \, dx = \cot x \, dx$. Thus,the integral becomes $I = \int \log t \, dt$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \log t$ and $dv = dt$: $I = t \log t - \int t \cdot \frac{1}{t} \, dt$. $I = t \log t - \int 1 \, dt$. $I = t \log t - t + c = t(\log t - 1) + c$. Substituting back $t = \log (\sin x)$: $I = \log (\sin x) [\log (\log (\sin x)) - 1] + c$.

$\int \cot x \cdot \log [\log (\sin x)] d x=$

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