A) 1
B) 2
C) 0
D) \[\pi \]
Correct Answer: C
Solution :
[c] Use \[\int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}}\] \[\int_{0}^{\pi }{\frac{\sin 2\,\,mx}{\sin \,\,x}dx=\int_{0}^{\pi }{\frac{\sin (2m\pi -2mx)}{\sin (\pi -x)}}dx}\] \[=\int_{0}^{\pi }{\frac{-\sin 2mx}{\sin x}dx=-I\Rightarrow 2I=0\Rightarrow I=0}\]
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