12) ∫π/6^π/3 e^tan x sec²x dx

Explanation:

Step1: Use substitution

Let $u = \tan x$, then $du=\sec^{2}x dx$. When $x = \frac{\pi}{6}$, $u=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$, when $x=\frac{\pi}{3}$, $u = \tan\frac{\pi}{3}=\sqrt{3}$. The integral becomes $\int_{\frac{\sqrt{3}}{3}}^{\sqrt{3}}e^{u}du$.

Step2: Integrate $e^{u}$

The antiderivative of $e^{u}$ is $e^{u}$. So, $\int_{\frac{\sqrt{3}}{3}}^{\sqrt{3}}e^{u}du=\left[e^{u}\right]_{\frac{\sqrt{3}}{3}}^{\sqrt{3}}$.

Step3: Evaluate the definite - integral

$e^{\sqrt{3}}-e^{\frac{\sqrt{3}}{3}}$.

Answer:

$e^{\sqrt{3}}-e^{\frac{\sqrt{3}}{3}}$