MCQ
The value of $\int_0^{2\pi } {{{\cos }^{99}}x\,dx} $ is
  • A
    $1$
  • B
    $ - 1$
  • C
    $99$
  • $0$

Answer

Correct option: D.
$0$
d
(d) 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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