MCQ
The value of $\int_{\,0}^{\,\pi /2} {\frac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\frac{\pi }{2}\,\, - \,\,x} \right)}^2}}}}}dx} $ is
- ✓$\pi /4$
- B$\pi /2$
- C${e^{{\pi ^2}/16}}$
- D${e^{{\pi ^2}/4}}$
$I = \int_0^{\pi /2} {\frac{{{e^{{{\left( {\frac{\pi }{2} - x} \right)}^2}}}\,\,\,\,\,\,dx}}{{{e^{{{\left( {\frac{\pi }{2}\, - x} \right)}^2}}} + {e^{{x^2}}}}}} $
$\left[ \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right]$
$ \Rightarrow 2I = \int_0^{\pi /2} {1dx = (x)_0^{\pi /2}} $
$ \Rightarrow I = \frac{\pi }{4}$.
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