# $\int_{\pi\over2}^{5\pi\over2}{e^{\tan^{-1}(\sin x)}\over e^{\tan^{-1}(\sin x)}+e^{\tan^{-1}(\cos x)}}dx$

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.

• Hint: $$\frac{e^{\tan^{-1}(\sin(x+\pi))}}{e^{\tan^{-1}(\sin(x+\pi))} + e^{\tan^{-1}(\cos(x+\pi))}} = \frac{e^{-\tan^{-1}(\sin(x))}}{e^{-\tan^{-1}(\sin(x))} + e^{-\tan^{-1}(\cos(x))}} = \frac{e^{\tan^{-1}(\cos(x))}}{e^{\tan^{-1}(\cos(x))} + e^{\tan^{-1}(\sin(x))}}$$ – achille hui May 8 '14 at 14:34
• Duplicate of: math.stackexchange.com/questions/781129/… – Pranav Arora May 8 '14 at 17:24

$$I:=\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\frac{e^{\tan^{-1}{(\sin{x})}}}{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}\,dx$$

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,

$$2I=\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\frac{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}{e^{\tan^{-1}{(\sin{x})}}+e^{\tan^{-1}{(\cos{x})}}}\,dx=\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}1\,dx=2\pi\\ \implies I=\pi.$$

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