MHB Definite Integral challenge #2

AI Thread Summary
The integral challenge involves evaluating the definite integral from π/2 to 5π/2 of the function e^(arctan(sin x)) divided by the sum of e^(arctan(sin x)) and e^(arctan(cos x)). The solution is approached by breaking the integral into two parts, I1 and I2, and applying symmetry transformations. Through these transformations, it is determined that I1 equals 5π/4 and I2 equals π/4. The final result of the integral is calculated to be π. The discussion highlights the elegance of the solution and the collaborative effort in solving the problem.
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Evaluate:
$$\Large \int_{\pi/2}^{5\pi/2} \frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}$$
 
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Pranav said:
Evaluate:
$$\Large \int_{\pi/2}^{5\pi/2} \frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}$$

$$\begin{aligned} \Large I & = \Large \int_{\pi/2}^{5\pi/2} \frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}\;{dx} \\& = \Large \int_{0}^{5\pi/2} \frac{e^{\arctan(\sin x)}\;{dx}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}-\int_{0}^{\pi/2} \frac{e^{\arctan(\sin x)}\;{dx}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}} \\& \Large= I_{1}-I_{2} \end{aligned}$$

Let $\displaystyle x \mapsto \frac{5\pi}{2}-x$ then $\displaystyle 2I_1 = \frac{5\pi}{2}$ thus $\displaystyle I_1 = \frac{5\pi}{4}$. Let $\displaystyle x \mapsto \frac{\pi}{2}-x$ then $\displaystyle 2I_2 = \frac{\pi}{2}$ thus $\displaystyle I_2 = \frac{\pi}{4}$. Thus $\displaystyle I = I_1-I_2 = \frac{5\pi}{4}-\frac{\pi}{4} = \pi$.
 
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Prometheus said:
$$\begin{aligned} \Large I & = \Large \int_{\pi/2}^{5\pi/2} \frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}\;{dx} \\& = \Large \int_{0}^{5\pi/2} \frac{e^{\arctan(\sin x)}\;{dx}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}-\int_{0}^{\pi/2} \frac{e^{\arctan(\sin x)}\;{dx}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}} \\& \Large= I_{1}-I_{2} \end{aligned}$$

Let $\displaystyle x \mapsto \frac{5\pi}{2}-x$ then $\displaystyle 2I_1 = \frac{5\pi}{2}$ thus $\displaystyle I_1 = \frac{5\pi}{4}$. Let $\displaystyle x \mapsto \frac{\pi}{2}-x$ then $\displaystyle 2I_2 = \frac{\pi}{2}$ thus $\displaystyle I_2 = \frac{\pi}{4}$. Thus $\displaystyle I = I_1-I_2 = \frac{5\pi}{4}-\frac{\pi}{4} = \pi$.

Brilliant Prometheus! :cool:

Thanks for your participation. :)
 
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