5.3 The Fundamental Theorem of Calculus/11

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

${\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{x}^{\pi }{\sqrt {1+sec(t)}}\cdot dt\right]=0\cdot {\sqrt {1+sec(\pi )}}-1\cdot {\sqrt {1+sec(x)}}=-{\sqrt {1+sec(x)}}$

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