MHT CET · Maths · Indefinite Integration
\(\int \frac{\log (\cot x)}{\sin 2 x} \mathrm{~d} x=\)
- 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.
- D \(\frac{-1}{4}(\log (\cot x))^2+c\), where \(c\) is constant of integration.
Answer & Solution
Correct Answer
(D) \(\frac{-1}{4}(\log (\cot x))^2+c\), where \(c\) is constant of integration.
Step-by-step Solution
Detailed explanation
Let \(\mathrm{I}=\int \frac{\log (\cot x)}{\sin 2 x} \mathrm{~d} x\)
Put \(\log (\cot x)=\mathrm{t}\)
\(\Rightarrow\left[\frac{1}{\cot x} \cdot\left(-\operatorname{cosec}^2 x\right)\right] \mathrm{d} x=\mathrm{dt} \)
\( \Rightarrow\left[\frac{-\sin x}{\sin ^2 x \cdot \cos x}\right] \mathrm{d} x=\mathrm{dt} \)
\( \Rightarrow \frac{\mathrm{d} x}{\sin 2 x}=\frac{-\mathrm{dt}}{2} \)
\( \therefore \mathrm{I} =\int \mathrm{t} \cdot\left(-\frac{\mathrm{dt}}{2}\right) \)
\( =\frac{-1}{4} \mathrm{t}^2+\mathrm{c} \)
\( =\frac{-1}{4}[\log (\cot x)]^2+\mathrm{c}\)
Put \(\log (\cot x)=\mathrm{t}\)
\(\Rightarrow\left[\frac{1}{\cot x} \cdot\left(-\operatorname{cosec}^2 x\right)\right] \mathrm{d} x=\mathrm{dt} \)
\( \Rightarrow\left[\frac{-\sin x}{\sin ^2 x \cdot \cos x}\right] \mathrm{d} x=\mathrm{dt} \)
\( \Rightarrow \frac{\mathrm{d} x}{\sin 2 x}=\frac{-\mathrm{dt}}{2} \)
\( \therefore \mathrm{I} =\int \mathrm{t} \cdot\left(-\frac{\mathrm{dt}}{2}\right) \)
\( =\frac{-1}{4} \mathrm{t}^2+\mathrm{c} \)
\( =\frac{-1}{4}[\log (\cot x)]^2+\mathrm{c}\)
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