_ ^ x ;dx 1 - x x = - Vidyadip

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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