Definite Integral

\hspace{-16}$Let $\bf{f(x)={ \frac{e^{tan^{-1}(sinx)}}{e^{tan^{-1}(sinx) }+ e^{tan^{-1}(cosx)}}} }$\\\\\\ and $\bf{f\left(\frac{\pi}{2}-x\right)={ \frac{e^{tan^{-1}(cosx)}}{e^{tan^{-1}(cos x) }+ e^{tan^{-1}(sin x)}}} }$\\\\\\ So $\bf{f(x)+f\left(\frac{\pi}{2}-x\right)=1}$\\\\\\ Now Integrate both side w.r.to $\bf{x}$ and Putting Limit.....\\\\\\ $\bf{\int_{\pi /2}^{5\pi /2}f(x)dx+\int_{\pi /2}^{5\pi /2}f\left(\frac{\pi}{2}-x\right)dx = \int_{\pi /2}^{5\pi /2}1.dx=2\pi}$\\\\\\ Now Put $\bf{\frac{\pi}{2}-x=t\Leftrightarrow dx=-dt}$ and Changing Limit\\\\\\ $\bf{\int_{\pi /2}^{5\pi /2}f(x)dx-\int_{0}^{2\pi}f(t)dt=2\pi}$\\\\\\ Using Definite Integral Property $\bf{\int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx}$\\\\\\ and $\bf{\int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx}$\\\\\\ and If $\bf{f(x)}$ is Periodic With Period $\bf{T>0\;,}$\\\\\\ Then $\bf{\int_{a}^{a+T}f(x)dx = \int_{0}^{T}f(x)dx}$\\\\\\

\hspace{-16}$So $\bf{\int_{\pi /2}^{5\pi /2}f(x)dx+\int_{0}^{2\pi}f(x)dx = 2\pi}$\\\\\\ So $\bf{\int_{\pi /2}^{5\pi /2}f(x)dx+\int_{\pi /2}^{5\pi /2}f(x)dx=2\pi}$\\\\\\ So $\boxed{\boxed{\bf{\int_{\pi /2}^{5\pi /2}f(x)dx=\pi}}}$

Yes. Very nice..