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Definite Integral challenge #2

Pranav

Well-known member
Nov 4, 2013
428
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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Prometheus

Member
Mar 25, 2013
11
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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Pranav

Well-known member
Nov 4, 2013
428
$$\begin{aligned} \Large I & = \Large \int_{\pi/2}^{5\pi/2} \frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}\;{dx} \\& = \Large \int_{0}^{5\pi/2} \frac{e^{\arctan(\sin x)}\;{dx}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}-\int_{0}^{\pi/2} \frac{e^{\arctan(\sin x)}\;{dx}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}} \\& \Large= I_{1}-I_{2} \end{aligned}$$

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:

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