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Câu hỏi: Cho $I=\int\limits_{1}^{2}{f\left( x \right)dx}=2$. Giá trị của $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin xf\left( \sqrt{3\cos x+1} \right)}{\sqrt{3\cos x+1}}dx}$ bằng
A. 2.
B. $-\dfrac{4}{3}$.
C. $\dfrac{4}{3}$.
D. 2.
A. 2.
B. $-\dfrac{4}{3}$.
C. $\dfrac{4}{3}$.
D. 2.
Đặt $t=\sqrt{3\cos x+1}\Rightarrow {{t}^{2}}=3\cos x+1\Rightarrow 2tdt=-3\sin xdx$.
Đối cận: $\left| \begin{aligned}
& x=0\Rightarrow t=2 \\
& x=\dfrac{\pi }{2}\Rightarrow t=1 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin xf\left( \sqrt{3\cos x+1} \right)}{\sqrt{3\cos x+1}}dx}=\int\limits_{2}^{1}{\dfrac{-2f\left( t \right)}{3}.\dfrac{tdt}{t}}=\dfrac{2}{3}\int\limits_{1}^{2}{f\left( t \right)dt}=\dfrac{2}{3}\int\limits_{1}^{2}{f\left( x \right)dx}=\dfrac{4}{3}$.
Đối cận: $\left| \begin{aligned}
& x=0\Rightarrow t=2 \\
& x=\dfrac{\pi }{2}\Rightarrow t=1 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin xf\left( \sqrt{3\cos x+1} \right)}{\sqrt{3\cos x+1}}dx}=\int\limits_{2}^{1}{\dfrac{-2f\left( t \right)}{3}.\dfrac{tdt}{t}}=\dfrac{2}{3}\int\limits_{1}^{2}{f\left( t \right)dt}=\dfrac{2}{3}\int\limits_{1}^{2}{f\left( x \right)dx}=\dfrac{4}{3}$.
Đáp án C.