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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( \dfrac{\pi }{2} \right)=1$ và ${f}'\left( x \right)=\cos x\left( 6{{\sin }^{2}}x-1 \right),\forall x\in \mathbb{R}$. Tính $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}$.
A. $\dfrac{5}{3}$.
B. $-\dfrac{5}{3}$.
C. $-\dfrac{2}{3}$.
D. $\dfrac{1}{3}$.
A. $\dfrac{5}{3}$.
B. $-\dfrac{5}{3}$.
C. $-\dfrac{2}{3}$.
D. $\dfrac{1}{3}$.
Ta có $f(x)=\int \cos x\left(6 \sin ^{2} x-1\right) \mathrm{d} x=2 \sin ^{3} x-\sin x+C$.
Từ $f\left(\dfrac{\pi}{2}\right)=1$ suy ra $C=0$.
Vậy $\int_{0}^{\dfrac{\pi}{2}} f(x) \mathrm{d} x=\int_{0}^{\dfrac{\pi}{2}}\left(2 \sin ^{3} x-\sin x\right) \mathrm{d} x=\dfrac{1}{3}$.
Từ $f\left(\dfrac{\pi}{2}\right)=1$ suy ra $C=0$.
Vậy $\int_{0}^{\dfrac{\pi}{2}} f(x) \mathrm{d} x=\int_{0}^{\dfrac{\pi}{2}}\left(2 \sin ^{3} x-\sin x\right) \mathrm{d} x=\dfrac{1}{3}$.
Đáp án D.