MCQ
$\int_{\, - \,\pi }^{\,\pi } {\frac{{{{\cos }^2}x}}{{1 + {a^x}}}dx,\,a > 0,} = . .$
- A$\pi $
- B$a\pi $
- ✓$\frac{\pi }{2}$
- D$2\pi $
$ = \int_{\, - \pi }^{\,\pi } {\frac{{{{\cos }^2}x}}{{1 + {a^{ - x}}}}} \,dx$
$ \Rightarrow I + I = \int_{\, - \pi }^{\,\pi } {{{\cos }^2}x\left( {\frac{1}{{1 + {a^x}}} + \frac{1}{{1 + {a^{ - x}}}}} \right)\,dx} $
$ = \int_{\, - \pi }^{\,\pi } {{{\cos }^2}x\,dx} $
==> $2I$ $ = 2\int_0^\pi {{{\cos }^2}x.\,dx = } \int_0^\pi {(1 + \cos 2x)dx} $
==> $2I$ $ = [x]_0^\pi + \left[ {\frac{{\sin 2x}}{2}} \right]_0^\pi $
$ \Rightarrow 2I = \pi \,\,\, \Rightarrow I = \frac{\pi }{2}$.
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