int(log (sin x)+x cot x) d x= | vedclass

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Question

(A) Let $I = \int (\log(\sin x) + x \cot x) dx$. We can split the integral as $I = \int \log(\sin x) dx + \int x \cot x dx$. Consider the first part $

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

Vedclass · Answer

(A) Let $I = \int (\log(\sin x) + x \cot x) dx$. We can split the integral as $I = \int \log(\sin x) dx + \int x \cot x dx$. Consider the first part $I_1 = \int \log(\sin x) dx$. Using integration by parts, let $u = \log(\sin x)$ and $dv = dx$. Then $du = \frac{1}{\sin x} \cdot \cos x dx = \cot x dx$ and $v = x$. Applying the integration by parts formula $\int u dv = uv - \int v du$, we get: $I_1 = x \log(\sin x) - \int x \cot x dx$. Substituting this back into the expression for $I$: $I = (x \log(\sin x) - \int x \cot x dx) + \int x \cot x dx + c$. $I = x \log(\sin x) + c$.

$\int(\log (\sin x)+x \cot x) d x=$

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