Question
$\int_0^{2\pi } {{{\cos }^{99}}x\,dx} =$
तब $I=2\int_{0}^{\pi }{{{\cos }^{99}}x\,dx,\,\,\,\{\because {{\cos }^{99}}(2\pi -x)={{\cos }^{99}}x\}}$
अब $\int_{0}^{\pi }{{{\cos }^{99}}x\,dx\,=0,\,\,\{\because {{\cos }^{99}}(\pi -x)=-{{\cos }^{99}}x\}}$
$\therefore \,\,I = 2 \times 0 = 0$.
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