int frac{x , dx}{1 - x cot x} = | vedclass

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Question

(C) We have the integral $I = \int \frac{x \, dx}{1 - x \cot x}$.Substitute $\cot x = \frac{\cos x}{\sin x}$:$I = \int \frac{x \, dx}{1 - x \frac{\cos

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We have the integral $I = \int \frac{x \, dx}{1 - x \cot x}$.Substitute $\cot x = \frac{\cos x}{\sin x}$:$I = \int \frac{x \, dx}{1 - x \frac{\cos x}{\sin x}} = \int \frac{x \sin x}{\sin x - x \cos x} \, dx$.Let $t = \sin x - x \cos x$.Differentiating with respect to $x$:$dt = (\cos x - (x(-\sin x) + \cos x)) \, dx = (\cos x + x \sin x - \cos x) \, dx = x \sin x \, dx$.Substituting into the integral:$I = \int \frac{dt}{t} = \log |t| + c = \log |\sin x - x \cos x| + c$.

$\int \frac{x \, dx}{1 - x \cot x} = $

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