\int_{-\pi /2}^{\pi /2}{\frac{cosx}{1+e^{x}}}dx
for practice.please post the solution too.
ans------->1
29Using the property of Definite integrals we have a∫b f(x) = a∫b f(a + b - x)
\int_{-\pi /2}^{\pi /2}{\frac{cosx}{1+ e^{x}}} = \int_{-\pi /2}^{\pi /2}{\frac{cosx}{1+ e^{-x}}} = \int_{-\pi /2}^{\pi /2}{\frac{e^{x}cosx}{1+ e^{x}}}
2I = \int_{-\pi /2}^{\pi /2}{cosx}dx = 2
so I = 1...
1\int_{-a}^{a}{f(x)}dx=\int_{0}^{a}[{f(x)+f(-x)]}
I=\int_{0}^{\pi /2}{cosx/(1+e^{x})+cos(-x)/)1+e^{-x)}}
I=\int_{0}^{\pi /2}{cosx}
=1.
1thnk s u guys.
posted it for practice and it seems u r all geniuses
