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The value of 0πsinnxcos2m+1xdx is 

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a
(2m+1)!(n!)2
b
(2m+1)!n!
c
∫0π cos2m−1⁡xdx
d
none of these

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detailed solution

Correct option is C

I=∫0π sinn⁡xcos2m+1⁡xdx   =∫0π sinn⁡(π−x)cos⁡2m+1(π−x)dx =−∫0π sinn⁡xcos⁡2m+1xdx=−ISo 2I=0 which implies that I=0 Also, ∫0π cos2m−1⁡xdx=0by a similar reasoning.


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