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If mn,m,nN then the value of 02πcosmx cosnxdx is

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detailed solution

Correct option is A

∫02π cos⁡mxcos⁡nxdx=12∫02π [cos⁡(m+n)x+cos⁡(m−n)x]dx=121m+nsin⁡(m+n)x+1m−nsin⁡(m−n)x02π=0


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