A) \[-1\]
B) 0
C) 1
D) \[\pi \]
Correct Answer: B
Solution :
Let \[f(x)=\int_{0}^{\pi }{{{e}^{{{\sin }^{2}}x}}{{\cos }^{3}}(2n+1)x.dx}\] Since \[\cos (2n+1)(\pi -x)=\cos [(2n+1)\pi -(2n+1)x]\] \[=-\cos (2n+1)x\]and \[{{\sin }^{2}}(\pi -x)={{\sin }^{2}}x\] Hence by the property of definite integral, \[\int_{0}^{\pi }{{{e}^{{{\sin }^{2}}x}}{{\cos }^{3}}(2n+1)x\,dx=0}\], \[[f(2a-x)=-f(x)]\].You need to login to perform this action.
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