A) \[\frac{(2m+1)!}{n!}\]
B) \[\frac{(2m+1)!}{{{n}^{2}}}\]
C) \[\int_{0}^{\pi }{{{\cos }^{2m+!}}}xdx\]
D) none of these
Correct Answer: D
Solution :
\[I=\int_{0}^{\pi }{{{\sin }^{n}}}x{{\cos }^{2m+1}}xdx\] \[=\int_{0}^{\pi }{{{\sin }^{n}}}(\pi -x).{{\cos }^{2m+1}}(\pi -x)dx\] \[=-\int_{0}^{\pi }{{{\sin }^{n}}}x{{\cos }^{2m+1}}xdx=-I\Rightarrow i=0\] Similarly, one can show that \[\int_{0}^{\pi }{{{\cos }^{2m+1}}}xdx=0\]
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