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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{x\;dx}}{{1 - x\cot x}}} = $
  • A
    $\log (\cos x - x\sin x) + c$
  • B
    $\log (x\sin x - \cos x) + c$
  • ✓
    $\log (\sin x - x\cos x) + c$
  • D
    None of these
✓

Answer

Correct option: C.
$\log (\sin x - x\cos x) + c$
c
(c)$\int_{}^{} {\frac{{x\,dx}}{{1 - x\cot x}}} = \int_{}^{} {\frac{{x\,dx}}{{1 - x\frac{{\cos x}}{{\sin x}}}}} = \int_{}^{} {\frac{{x\sin x}}{{\sin x - x\cos x}}\,dx} $
$ = \int_{}^{} {\frac{{dt}}{t}} = \log t = \log (\sin x - x\cos x) + c.$
$\{$Putting $\sin x - x\cos x = t$,$\}$
$⇒$  $[\cos x - ( - x\sin x + \cos x)]\,dx = dt \Rightarrow x\sin x\,dx = dt\} $

$ = \int_{}^{} {\frac{{dt}}{t}}  = \log t = \log (\sin x - x\cos x) + c.$

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