Question

What is the lim as x approaches pi [integral 1+tan t from pi to x]/(pi sin x)

Answer 1

Apply l'Hopital's rule:

$\displaystyle\lim_{x\to\pi}\frac{\displaystyle\int_\pi^x (1+\tan(t))\,\mathrm dt}{\pi\sin(x)}=\lim_{x\to\pi}\frac{1+\tan(x)}{\pi\cos(x)}=\frac{1+\tan(\pi)}{\pi\cos(\pi)}=\boxed{-\frac1\pi}$

where

$\displaystyle\frac{\mathrm d}{\mathrm dx}\left[\int_\pi^x(1+\tan(t))\,\mathrm dt\right]=1+\tan(x)$

follows from the fundamental theorem of calculus.