A) \[x\cot \frac{x}{2}+c\]
B) \[-x\cot \frac{x}{2}+c\]
C) \[\cot \frac{x}{2}+c\]
D) None of these
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{x-\sin x}{1-\cos x}\,dx}=\int_{{}}^{{}}{\frac{x}{1-\cos x}\,dx}-\int_{{}}^{{}}{\frac{\sin x}{1-\cos x}\,dx}\] \[=\frac{1}{2}\int_{{}}^{{}}{x\,\text{cose}{{\text{c}}^{\text{2}}}\left( \frac{x}{2} \right)\,dx}-\int_{{}}^{{}}{\frac{2\sin (x/2)\cos (x/2)}{2{{\sin }^{2}}(x/2)}\,dx}\] \[=\frac{1}{2}\int_{{}}^{{}}{x\,\text{cose}{{\text{c}}^{\text{2}}}\left( \frac{x}{2} \right)\,dx}-\int_{{}}^{{}}{\cot \left( \frac{x}{2} \right)\,dx}\]\[=-x\cot \left( \frac{x}{2} \right)+c\].
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