5.3 The Fundamental Theorem of Calculus/11

$g(x)=\int _{x}^{\pi }{\sqrt {1+sec(t)}}\cdot dt$

Failed to parse (unknown function "\x"): {\displaystyle \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\x^{\pi}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}}

${\text{Therefore, }}g'(x)=-{\sqrt {1+sec(x)}}$