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Câu hỏi: Tìm nguyên hàm $F\left( x \right)=\int\limits_{{}}^{{}}{{{\sin }^{2}}2xdx}$
A. $F\left( x \right)=\dfrac{1}{2}x-\dfrac{1}{8}\cos 4x+C$.
B. $F\left( x \right)=\dfrac{1}{2}x-\dfrac{1}{8}\sin 4x+C$.
C. $F\left( x \right)=\dfrac{1}{2}x-\dfrac{1}{8}\sin 4x$.
D. $F\left( x \right)=\dfrac{1}{2}x+\dfrac{1}{8}\sin 4x+C$.
A. $F\left( x \right)=\dfrac{1}{2}x-\dfrac{1}{8}\cos 4x+C$.
B. $F\left( x \right)=\dfrac{1}{2}x-\dfrac{1}{8}\sin 4x+C$.
C. $F\left( x \right)=\dfrac{1}{2}x-\dfrac{1}{8}\sin 4x$.
D. $F\left( x \right)=\dfrac{1}{2}x+\dfrac{1}{8}\sin 4x+C$.
Ta có $F\left( x \right)=\int\limits_{{}}^{{}}{{{\sin }^{2}}2xdx}=\int\limits_{{}}^{{}}{\dfrac{1-\cos 4x}{2}dx}=\dfrac{1}{2}\int\limits_{{}}^{{}}{1dx}-\dfrac{1}{2}\int\limits_{{}}^{{}}{\cos 4xdx}=\dfrac{1}{2}x-\dfrac{1}{8}\sin 4x+C$.
Đáp án B.