Q99 / 150 · MHT CET 2023 (11 May Shift 1) · Mathematics

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

  • A
    $-\log (\cot x)^2+c$, where $c$ is constant of integration.
  • B
    $2(\log (\cot x))^2+c$, where $\mathrm{c}$ is constant of integration.
  • C
    $\frac{-1}{4}(\log (\sin x))^2+c$, where $\mathrm{c}$ is constant of integration.
  • Correct answer
    D
    $\frac{-1}{4}(\log (\cot x))^2+c$, where $c$ is constant of integration.

Solution

Let $\mathrm{I}=\int \frac{\log (\cot x)}{\sin 2 x} \mathrm{~d} x$ Put $\log (\cot x)=\mathrm{t}$ $$ \begin{aligned} & \Rightarrow\left[\frac{1}{\cot x} \cdot\left(-\operatorname{cosec}^2 x\right)\right] \mathrm{d} x=\mathrm{dt} \\ & \Rightarrow\left[\frac{-\sin x}{\sin ^2 x \cdot \cos x}\right] \mathrm{d} x=\mathrm{dt} \\ & \Rightarrow \frac{\mathrm{d} x}{\sin 2 x}=\frac{-\mathrm{dt}}{2} \\ \therefore \quad \mathrm{I} & =\int \mathrm{t} \cdot\left(-\frac{\mathrm{dt}}{2}\right) \\ & =\frac{-1}{4} \mathrm{t}^2+\mathrm{c} \\ & =\frac{-1}{4}[\log (\cot x)]^2+\mathrm{c} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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