Question

If

Answer 1

It's easy to show that $7\tan(4x)$ is strictly increasing on $x\in\left[0,\frac\pi8\right]$. This means

$M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3$

and

$m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7$

Then the integral is bounded by

$\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)$

$\implies \displaystyle \boxed{\frac{7\pi}{48}} \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le \boxed{\frac{7\sqrt3\,\pi}{48}}$