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Câu hỏi: Cho hàm số $f(x)=\left\{\begin{array}{l}2 x-2 \quad \text { khi } x \leq 0 \\ x^2+4 x-2 \text { khi } x>0\end{array}\right.$. Tích phân $I=\int_0^\pi \sin 2 x . f(\cos x) \mathrm{d} x$ bằng
A. $I=\dfrac{7}{6}$.
B. $I=\dfrac{9}{2}$.
C. $I=-\dfrac{9}{2}$.
D. $I=-\dfrac{7}{6}$.
A. $I=\dfrac{7}{6}$.
B. $I=\dfrac{9}{2}$.
C. $I=-\dfrac{9}{2}$.
D. $I=-\dfrac{7}{6}$.
Do $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)=-2$ nên hàm số $f(x)$ liên tục tại điểm $x=0$.
Đặt $t=\cos x \Rightarrow \mathrm{d} t=-\sin x \mathrm{~d} x$.
Đổi cận: $x=0 \Rightarrow t=1 ; x=\pi \Rightarrow t=-1$.
Ta có: $\int_0^\pi \sin 2 x . f(\cos x) \mathrm{d} x=\int_0^\pi 2 \sin x . \cos x . f(\cos x) \mathrm{d} x=-\int_1^{-1} 2 t . f(t) \mathrm{d} t=2 \int_{-1}^1 t . f(t) \mathrm{d} t$ $=2 \int_{-1}^0 x \cdot f(x) \mathrm{d} x+2 \int_0^1 x \cdot f(x) \mathrm{d} x=2 \int_0^1 x\left(x^2+4 x-2\right) \mathrm{d} x+2 \int_{-1}^0 x \cdot(2 x-2) \mathrm{d} x$ $=\left.2\left(\dfrac{x^4}{4}+\dfrac{4 x^3}{3}-x^2\right)\right|_0 ^1+\left.4 \cdot\left(\dfrac{x^3}{3}-\dfrac{x^2}{2}\right)\right|_{-1} ^0=\dfrac{7}{6}+\dfrac{10}{3}=\dfrac{9}{2}$.
Đặt $t=\cos x \Rightarrow \mathrm{d} t=-\sin x \mathrm{~d} x$.
Đổi cận: $x=0 \Rightarrow t=1 ; x=\pi \Rightarrow t=-1$.
Ta có: $\int_0^\pi \sin 2 x . f(\cos x) \mathrm{d} x=\int_0^\pi 2 \sin x . \cos x . f(\cos x) \mathrm{d} x=-\int_1^{-1} 2 t . f(t) \mathrm{d} t=2 \int_{-1}^1 t . f(t) \mathrm{d} t$ $=2 \int_{-1}^0 x \cdot f(x) \mathrm{d} x+2 \int_0^1 x \cdot f(x) \mathrm{d} x=2 \int_0^1 x\left(x^2+4 x-2\right) \mathrm{d} x+2 \int_{-1}^0 x \cdot(2 x-2) \mathrm{d} x$ $=\left.2\left(\dfrac{x^4}{4}+\dfrac{4 x^3}{3}-x^2\right)\right|_0 ^1+\left.4 \cdot\left(\dfrac{x^3}{3}-\dfrac{x^2}{2}\right)\right|_{-1} ^0=\dfrac{7}{6}+\dfrac{10}{3}=\dfrac{9}{2}$.
Đáp án B.