g ( x ) = ∫ x π 1 + s e c ( t ) ⋅ d t {\displaystyle g(x)=\int _{x}^{\pi }{\sqrt {1+sec(t)}}\cdot dt} d d x [ g ( x ) ] = d d x [ ∫ x π 1 + s e c ( t ) ⋅ d t ] = 0 ⋅ 1 + s e c ( π ) − 1 ⋅ 1 + s e c ( x ) = − 1 + s e c ( 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)}}} Therefore, g ′ ( x ) = − 1 + s e c ( x ) {\displaystyle {\text{Therefore, }}g'(x)=-{\sqrt {1+sec(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)}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/61cd7fe7ef01b603b4b8f95f5036e31f26c4e84a)
