A) \[\log (\cos x-x\sin x)+c\]
B) \[\log (x\sin x-\cos x)+c\]
C) \[\log (\sin x-x\cos x)+c\]
D) None of these
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{x\,dx}{1-x\cot x}}=\int_{{}}^{{}}{\frac{x\,dx}{1-x\frac{\cos x}{\sin x}}}=\int_{{}}^{{}}{\frac{x\sin x}{\sin x-x\cos x}\,dx}\] \[=\int_{{}}^{{}}{\frac{dt}{t}}=\log t=\log (\sin x-x\cos x)+c.\] {Putting \[\sin x-x\cos x=t\], Þ \[[\cos x-(-x\sin x+\cos x)]\,dx=dt\Rightarrow x\sin x\,dx=dt\}\]
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