A) 1
B) \[-1\]
C) 99
D) 0
Correct Answer: D
Solution :
Let \[I=\int_{0}^{2\pi }{{{\cos }^{99}}x\,dx.}\] Then \[I=2\int_{0}^{\pi }{{{\cos }^{99}}x\,dx,\,\,\,\{\because {{\cos }^{99}}(2\pi -x)={{\cos }^{99}}x\}}\] Now, \[\int_{0}^{\pi }{{{\cos }^{99}}x\,dx\,=0,\,\,\{\because {{\cos }^{99}}(\pi -x)=-{{\cos }^{99}}x\}}\] \[\therefore \,\,I=2\times 0=0\].
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