int frac{log (cot x)}{sin 2 x} ,d x= | vedclass

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Question

(D) Let $I = \int \frac{\log (\cot x)}{\sin 2 x} \,d x$. Put $\log (\cot x) = t$. Differentiating both sides with respect to $x$: $\frac{1}{\cot x} \c

Vedclass · Accepted Answer

$\frac{-1}{4}(\log (\cot x))^2+c$, where $c$ is constant of integration.

Vedclass · Answer

(D) Let $I = \int \frac{\log (\cot x)}{\sin 2 x} \,d x$. Put $\log (\cot x) = t$. Differentiating both sides with respect to $x$: $\frac{1}{\cot x} \cdot (-\csc^2 x) \,d x = dt$. $\Rightarrow \frac{\sin x}{\cos x} \cdot (-\frac{1}{\sin^2 x}) \,d x = dt$. $\Rightarrow \frac{-1}{\sin x \cos x} \,d x = dt$. Since $\sin 2x = 2 \sin x \cos x$, we have $\sin x \cos x = \frac{\sin 2x}{2}$. Substituting this: $\frac{-1}{\frac{\sin 2x}{2}} \,d x = dt \Rightarrow \frac{-2}{\sin 2x} \,d x = dt \Rightarrow \frac{d x}{\sin 2x} = \frac{-dt}{2}$. Now, substitute these into the integral: $I = \int t \cdot (\frac{-dt}{2}) = -\frac{1}{2} \int t \,dt$. $I = -\frac{1}{2} \cdot \frac{t^2}{2} + c = -\frac{1}{4} t^2 + c$. Substituting $t = \log (\cot x)$ back: $I = -\frac{1}{4} [\log (\cot x)]^2 + c$.

$\int \frac{\log (\cot x)}{\sin 2 x} \,d x=$

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