1
$\begingroup$

$$\int_0^{2\pi} \frac{e^{|\sin x|}\cos(x)}{1+e^{\tan x}} \, dx$$

My try:

$$I=\int_0^\pi \frac{e^{\sin x}\cos(x)}{1+e^{\tan x}} dx+\int_\pi^{2\pi} \frac{e^{-\sin x}\cos(x)}{1+e^{\tan x}} dx$$

also

$$I=-\int_0^\pi \frac{e^{\sin x}\cos(x)}{1+e^{-\tan x}} \,dx-\int_\pi^{2\pi} \frac{e^{-\sin x}\cos(x)}{1+e^{-\tan x}} \, dx$$

using $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$

Now how to proceed ?

$\endgroup$
  • $\begingroup$ Did you try in the second term of integral make $x=y+\pi$, $0\leq y\leq \pi$? $\endgroup$ – Euler88 ... Aug 1 '15 at 20:08
  • $\begingroup$ No..does that help ? @Euler88... $\endgroup$ – user220382 Aug 1 '15 at 20:09
  • 1
    $\begingroup$ Yeah.. Finish the problem! Good Luck! $\endgroup$ – Euler88 ... Aug 1 '15 at 20:15
3
$\begingroup$

Let: $$ f(x) = \frac{e^{\left|\sin x\right|}\cos x}{1+e^{\tan{x}}}.$$ Since $\sin(\pi+x)=-\sin(x)$, $\cos(\pi+x)=-\cos(x)$ and $\tan(x+\pi)=\tan(x)$, by assuming that $f(x)$ is integrable over $(0,\pi)$ we have: $$ \int_{0}^{2\pi}f(x)\,dx = \int_{0}^{\pi}f(x)+f(x+\pi)\,dx = 0.$$

$\endgroup$
  • 1
    $\begingroup$ Now the real achievement would be finding the value of the integral 0 to $\pi$ ;) $\endgroup$ – Zach466920 Aug 1 '15 at 20:15

Your Answer

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