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