$\int_{\pi\over2}^{5\pi\over2}{e^{\tan^{-1}(\sin x)}\over e^{\tan^{-1}(\sin x)}+e^{\tan^{-1}(\cos x)}}dx$ Find the value of the integral
$$\int_{\pi/2}^{5\pi/2}{e^{\tan^{-1}(\sin x)}\over e^{\tan^{-1}(\sin x)}+e^{\tan^{-1}(\cos x)}}dx$$
I was trying to use the property $\int_a^bf(x)dx=\int_a^bf(a+b-x)$
However I am unable to evaluate it.
 A: Break the integral as:
 $$\int_0 ^{\frac{5\pi}{2}}f(x)dx-\int_0^{\frac{\pi}{2}}f(x)dx $$
This allows you to use that property and evaluate those two integrals independently as denominator remains same and on adding, numerator=denominator
A: $$I:=\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\frac{e^{\tan^{-1}{(\sin{x})}}}{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}\,dx$$
Building on achille hui's hint in the comments, substitute $x=u+\pi$ to find,
$$I=\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{e^{\tan^{-1}{(\cos{x})}}}{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}\,dx.$$
Since the integrand is periodic in $x$ with period $2\pi$, any integral over an interval of length $2\pi$ will have the same value (i.e., we can shift the limits of integration by an arbitrary amount. Hence,
$$I=\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{e^{\tan^{-1}{(\cos{x})}}}{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}\,dx=\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\frac{e^{\tan^{-1}{(\cos{x})}}}{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}\,dx.$$
Thus,
$$2I=\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\frac{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}\,dx=\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}1\,dx=2\pi\\
\implies I=\pi.$$
