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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và $\int\limits_{-1}^{1}{f\left( x \right)} \text{d}x=6$, $\int\limits_{\dfrac{\pi }{3}}^{\dfrac{2\pi }{3}}{f\left( 2\cos x \right)\sin x }\text{d}x$ bằng
A. $-3$.
B. $3$.
C. $12$.
D. $-12$.
A. $-3$.
B. $3$.
C. $12$.
D. $-12$.
$\int\limits_{\dfrac{\pi }{3}}^{\dfrac{2\pi }{3}}{f\left( 2\cos x \right)\sin x }\text{d}x$
Đặt $t=2\cos x\Rightarrow dt=-2\sin xdx\Rightarrow -\dfrac{1}{2}dt=\sin xdx$
Đổi cận:
$\begin{aligned}
& x=\dfrac{\pi }{3} \Rightarrow t=1 \\
& x=\dfrac{2\pi }{3}\Rightarrow t=-1 \\
\end{aligned}$
$\int\limits_{\dfrac{\pi }{3}}^{\dfrac{2\pi }{3}}{f\left( 2\cos x \right)\sin x }\text{d}x=\int\limits_{1}^{-1}{f\left( t \right)-\dfrac{1}{2}dt }=-\dfrac{1}{2}\int\limits_{1}^{-1}{f\left( t \right)dt }=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( t \right)dt }=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( x \right)dx=\dfrac{1}{2}.6 =3}$
Đặt $t=2\cos x\Rightarrow dt=-2\sin xdx\Rightarrow -\dfrac{1}{2}dt=\sin xdx$
Đổi cận:
$\begin{aligned}
& x=\dfrac{\pi }{3} \Rightarrow t=1 \\
& x=\dfrac{2\pi }{3}\Rightarrow t=-1 \\
\end{aligned}$
$\int\limits_{\dfrac{\pi }{3}}^{\dfrac{2\pi }{3}}{f\left( 2\cos x \right)\sin x }\text{d}x=\int\limits_{1}^{-1}{f\left( t \right)-\dfrac{1}{2}dt }=-\dfrac{1}{2}\int\limits_{1}^{-1}{f\left( t \right)dt }=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( t \right)dt }=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( x \right)dx=\dfrac{1}{2}.6 =3}$
Đáp án B.