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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of ∫ ( ( x))/( 2x)dx is equal to (where, C is the...

The value of ∫ln⁡(cot⁡x)sin⁡2xdx∫ \dfrac{\ln \left(\cot x\right)}{\sin 2x}dx is equal to (where, CC is the constant of integration)

Options

  1. A(ln⁡(cotx))22+C\dfrac{{\left(\ln \left(cot x\right)\right)}^{2}}{2} + C
  2. B(ln⁡(cotx))24+C\dfrac{{\left(\ln \left(cot x\right)\right)}^{2}}{4} + C
  3. C(ln⁡(cotx))26+C\dfrac{{\left(\ln \left(cot x\right)\right)}^{2}}{6} + C
  4. D−14(ln⁡(cotx))2+C- \dfrac{1}{4}{\left(\ln \left(cot x\right)\right)}^{2} + C

Correct answer

D. −14(ln⁡(cotx))2+C- \dfrac{1}{4}{\left(\ln \left(cot x\right)\right)}^{2} + C

Step-by-step solution

∫ln⁡(cotx)2sinxcosxdx=I∫ \dfrac{\ln \left(cot x\right)}{2sin xcos x}dx = Ι (let)
Put ln⁡(cotx)=t\ln \left(cot x\right) = t
⇒1cotx×cosec2xdx=−dt⇒ \dfrac{1}{cot x} \times {cosec}^{2} xdx = - dt
⇒dxcosxsinx=−dt⇒ \dfrac{dx}{cos xsin x} = - dt
So, I=12∫(−t)dt=−12(t22)+CΙ = \dfrac{1}{2}∫ \left(- t\right)dt = \dfrac{- 1}{2}\left(\dfrac{t^{2}}{2}\right) + C
=−14(ln⁡(cotx))2+C= - \dfrac{1}{4}{\left(\ln \left(cot x\right)\right)}^{2} + C
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