Câu hỏi: Cho
A.
B.
C. $$ ${F}'\left( x \right)=-\dfrac{1}{{{\sin }^{2}}x}D. \)"> {F}'\left( x \right)=\dfrac{1}{{{\sin }^{2}}x}
A.
B.
C. $$ ${F}'\left( x \right)=-\dfrac{1}{{{\sin }^{2}}x}
Có \)">\int{\dfrac{1}{{{\sin }^{2}}x}\text{d}x}=F\left( x \right)+C\Rightarrow {F}'\left( x \right)=\dfrac{1}{{{\sin }^{2}}x}$.
Đáp án D.