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Câu hỏi: Cho hàm số $f\left( x \right)=\sin x\text{cos}x$. Khẳng định nào dưới đây đúng?
A. $\int{f\left( x \right)}\text{d}x=\sin x+\text{cos}x+C$.
B. $\int{f\left( x \right)}\text{d}x=\dfrac{1}{2}{{\sin }^{2}}x+C$.
C. $\int{f\left( x \right)}\text{d}x={{\sin }^{2}}x+C$.
D. $\int{f\left( x \right)}\text{d}x=\dfrac{1}{2}\text{co}{{\text{s}}^{2}}x+C$.
A. $\int{f\left( x \right)}\text{d}x=\sin x+\text{cos}x+C$.
B. $\int{f\left( x \right)}\text{d}x=\dfrac{1}{2}{{\sin }^{2}}x+C$.
C. $\int{f\left( x \right)}\text{d}x={{\sin }^{2}}x+C$.
D. $\int{f\left( x \right)}\text{d}x=\dfrac{1}{2}\text{co}{{\text{s}}^{2}}x+C$.
Ta có: $f\left( x \right)=\sin x\text{cos}x=\dfrac{1}{2}\sin 2x\Rightarrow \int{f\left( x \right)}\text{d}x=\dfrac{1}{2}\int{\sin 2x}\text{d}x=-\dfrac{1}{4}\text{cos2}x+{{C}_{1}}$
$\Rightarrow \int{f\left( x \right)}\text{d}x=-\dfrac{1}{4}\left( 1-2{{\sin }^{2}}x \right)+{{C}_{1}}=\dfrac{1}{2}{{\sin }^{2}}x+C$.
$\Rightarrow \int{f\left( x \right)}\text{d}x=-\dfrac{1}{4}\left( 1-2{{\sin }^{2}}x \right)+{{C}_{1}}=\dfrac{1}{2}{{\sin }^{2}}x+C$.
Đáp án B.