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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)={{\sin }^{2}}x{{\cos }^{2}}x$ là:
A. $\dfrac{1}{4}x-\dfrac{1}{16}\sin 4x+C.$
B. $\dfrac{1}{8}x-\dfrac{1}{32}\sin 4x.$
C. $\dfrac{1}{8}x-\dfrac{1}{8}\sin 4x+C.$
D. $\dfrac{1}{4}x-\dfrac{1}{32}\sin 4x+C.$
A. $\dfrac{1}{4}x-\dfrac{1}{16}\sin 4x+C.$
B. $\dfrac{1}{8}x-\dfrac{1}{32}\sin 4x.$
C. $\dfrac{1}{8}x-\dfrac{1}{8}\sin 4x+C.$
D. $\dfrac{1}{4}x-\dfrac{1}{32}\sin 4x+C.$
Ta có: $f\left( x \right)={{\sin }^{2}}x{{\cos }^{2}}x=\dfrac{1}{4}{{\sin }^{2}}2x=\dfrac{1}{4}\left( \dfrac{1-\cos 4x}{2} \right)=\dfrac{1}{8}-\dfrac{1}{8}\cos 4x$.
Do đó $\int{f\left( x \right)dx}=\int{\left( \dfrac{1}{8}-\dfrac{1}{8}\cos 4x \right)dx}=\dfrac{1}{8}x-\dfrac{1}{8}.\dfrac{1}{4}.\sin 4x+C=\dfrac{1}{8}x-\dfrac{1}{32}\sin 4x+C$.
Do đó $\int{f\left( x \right)dx}=\int{\left( \dfrac{1}{8}-\dfrac{1}{8}\cos 4x \right)dx}=\dfrac{1}{8}x-\dfrac{1}{8}.\dfrac{1}{4}.\sin 4x+C=\dfrac{1}{8}x-\dfrac{1}{32}\sin 4x+C$.
Đáp án D.