4
$\begingroup$

The question is to evaluate $$\int_{\pi/2}^{5\pi/2} \frac{e^{\tan^{-1} \sin x}}{e^{\tan^{-1} \sin x}+e^{\tan^{-1} \cos x}}dx$$

I tried to take idea from the graph of $\tan^{-1} \tan x$ and rewrite the integral as $$\int_{\pi/2}^{5\pi/2} \frac{e^{\tan^{-1} \sin x}}{e^{\tan^{-1} \sin x}+e^{\tan^{-1} -\cos x}}dx$$.I couldn't proceed from here.Any ideas?Thanks.

$\endgroup$
4
$\begingroup$

Upon enforcing the substitution $x=\pi/2 -y$ we see that

$$\begin{align} I&=\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\sin(x))}}{e^{\arctan(\sin(x))}+e^{\arctan(\cos(x))}}\,dx\\\\ &=\int_0^{2\pi}\frac{e^{\arctan(\cos(y))}}{e^{\arctan(\cos(y))}+e^{\arctan(\sin(y))}}\,dy\tag 1 \end{align}$$

Noting that the integrand is $2\pi$-periodic, we can write

$$\int_0^{2\pi}\frac{e^{\arctan(\cos(y))}}{e^{\arctan(\cos(y))}+e^{\arctan(\sin(y))}}\,dy=\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\cos(y))}}{e^{\arctan(\cos(y))}+e^{\arctan(\sin(y))}}\,dy \tag 2$$

Using $(2)$ in $(1)$ we see that

$$I=\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\cos(x))}}{e^{\arctan(\sin(x))}+e^{\arctan(\cos(x))}}\,dx \tag 3$$

Therefore, by adding $(1)$ and $(3)$ and dividing by $2$ yields the coveted result

$$\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\sin(x))}}{e^{\arctan(\sin(x))}+e^{\arctan(\cos(x))}}\,dx=\pi$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.