Google
Câu hỏi: Cho hàm số $f(x)=\left\{\begin{array}{lll}2 x-1 & \text { khi } & x>0 \\ x^2-x-1 & \text { khi } & x \leq 0\end{array}\right.$. Tích phân $\int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin 2 x f^{\prime}(\sin x) \mathrm{d} x$ bằng
A. $\dfrac{11}{3}$.
B. $\dfrac{13}{3}$.
C. $\dfrac{5}{3}$.
D. $\dfrac{19}{3}$.
A. $\dfrac{11}{3}$.
B. $\dfrac{13}{3}$.
C. $\dfrac{5}{3}$.
D. $\dfrac{19}{3}$.
Ta có $I=\int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin 2 x f^{\prime}(\sin x) \mathrm{d} x=\int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} 2 \sin x f^{\prime}(\sin x) \cos x \mathrm{~d} x$ $=2 \int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin x \mathrm{~d}(f(\sin x))=\left.[2 \sin x f(\sin x)]\right|_{-\dfrac{\pi}{2}} ^{\dfrac{\pi}{2}}-2 \int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} f(\sin x) \cos x \mathrm{~d} x=2 f(1)+2 f(-1)-I_1$ Ta có: $f(1)=2 \cdot 1-1=1 ; \quad f(-1)=(-1)^2-(-1)-1=1$.
Tính $I_1=2 \int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} f(\sin x) \cos x \mathrm{~d} x$.
Đặt $\sin x=t$. Suy ra $I_1=2 \int_{-1}^1 f(t) \mathrm{d} t=2 \int_{-1}^0 f(t) \mathrm{d} t+2 \int_0^1 f(t) \mathrm{d} t=2 \int_{-1}^0\left(t^2-t-1\right) \mathrm{d} t+$ $2 \int_0^1(2 t-1) \mathrm{d} t=-\dfrac{1}{3}$.
Do vậy $I=2(1+1)-\left(-\dfrac{1}{3}\right)=\dfrac{13}{3}$.
Tính $I_1=2 \int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} f(\sin x) \cos x \mathrm{~d} x$.
Đặt $\sin x=t$. Suy ra $I_1=2 \int_{-1}^1 f(t) \mathrm{d} t=2 \int_{-1}^0 f(t) \mathrm{d} t+2 \int_0^1 f(t) \mathrm{d} t=2 \int_{-1}^0\left(t^2-t-1\right) \mathrm{d} t+$ $2 \int_0^1(2 t-1) \mathrm{d} t=-\dfrac{1}{3}$.
Do vậy $I=2(1+1)-\left(-\dfrac{1}{3}\right)=\dfrac{13}{3}$.
Đáp án B.