Integral Calculus Question 384
Question: $ \int_{{}}^{{}}{\frac{x-\sin x}{1-\cos x}dx=} $
[AISSE 1989]
Options:
A) $ x\cot \frac{x}{2}+c $
B) $ -x\cot \frac{x}{2}+c $
C) $ \cot \frac{x}{2}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{x-\sin x}{1-\cos x},dx}=\int_{{}}^{{}}{\frac{x}{1-\cos x},dx}-\int_{{}}^{{}}{\frac{\sin x}{1-\cos x},dx} $ $ =\frac{1}{2}\int_{{}}^{{}}{x,cose{c^{2}}( \frac{x}{2} ),dx}-\int_{{}}^{{}}{\frac{2\sin (x/2)\cos (x/2)}{2{{\sin }^{2}}(x/2)},dx} $ $ =\frac{1}{2}\int_{{}}^{{}}{x,cose{c^{2}}( \frac{x}{2} ),dx}-\int_{{}}^{{}}{\cot ( \frac{x}{2} ),dx} $ $ =-x\cot ( \frac{x}{2} )+c $ .
BETA
