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Câu hỏi: Cho hàm số $f(x)=\left\{\begin{array}{llc}3 x+2 x^3 & \text { khi } & x>-2 \\ 4-3 x & \text { khi } & x \leq-2\end{array}\right.$.
Tích phân $\int_{-\dfrac{\pi}{4}}^0 f(-1+2 \sin 2 x) \cdot \cos 2 x \cdot d x$ bằng
A. $\dfrac{1}{8}$.
B. $-\dfrac{1}{4}$.
C. $\dfrac{47}{8}$.
D. $-\dfrac{1}{8}$.
Tích phân $\int_{-\dfrac{\pi}{4}}^0 f(-1+2 \sin 2 x) \cdot \cos 2 x \cdot d x$ bằng
A. $\dfrac{1}{8}$.
B. $-\dfrac{1}{4}$.
C. $\dfrac{47}{8}$.
D. $-\dfrac{1}{8}$.
$
\begin{aligned}
& \text { Đặt } u=-1+2 \sin 2 x \Rightarrow d u=4 \cos 2 x \cdot d x \Rightarrow \cos 2 x \cdot d x=\dfrac{d u}{4} \text {. } \\
& \text { Với } x=-\dfrac{\pi}{4} \Rightarrow u=-3 . \\
& \text { Với } x=0 \stackrel{\Rightarrow}{\Rightarrow} u=-1 . \\
& \int_{-\dfrac{\pi}{4}}^0 f(-1+2 \sin 2 x) \cdot \cos 2 x \cdot d x=\int_{-3}^{-1} f(u) \cdot \dfrac{d u}{4}=\dfrac{1}{4}\left(\int_{-3}^{-2} f(u) \cdot d u+\int_{-2}^{-1} f(u) \cdot d u\right) \\
& =\dfrac{1}{4}\left(\int_{-3}^{-2}(4-3 u) \cdot d u+\int_{-2}^{-1}\left(3 u+2 u^3\right) \cdot d u\right)=\dfrac{1}{4}\left(\dfrac{23}{2}-12\right)=-\dfrac{1}{8} .
\end{aligned}
$
\begin{aligned}
& \text { Đặt } u=-1+2 \sin 2 x \Rightarrow d u=4 \cos 2 x \cdot d x \Rightarrow \cos 2 x \cdot d x=\dfrac{d u}{4} \text {. } \\
& \text { Với } x=-\dfrac{\pi}{4} \Rightarrow u=-3 . \\
& \text { Với } x=0 \stackrel{\Rightarrow}{\Rightarrow} u=-1 . \\
& \int_{-\dfrac{\pi}{4}}^0 f(-1+2 \sin 2 x) \cdot \cos 2 x \cdot d x=\int_{-3}^{-1} f(u) \cdot \dfrac{d u}{4}=\dfrac{1}{4}\left(\int_{-3}^{-2} f(u) \cdot d u+\int_{-2}^{-1} f(u) \cdot d u\right) \\
& =\dfrac{1}{4}\left(\int_{-3}^{-2}(4-3 u) \cdot d u+\int_{-2}^{-1}\left(3 u+2 u^3\right) \cdot d u\right)=\dfrac{1}{4}\left(\dfrac{23}{2}-12\right)=-\dfrac{1}{8} .
\end{aligned}
$
Đáp án D.