Integral Calculus Question 102
Question: $ \int_{{}}^{{}}{\frac{x\ dx}{1-x\cot x}}= $
Options:
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
Show Answer
Answer:
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} $
BETA
