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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. 2
C. $-\dfrac{4}{3}$
D. $\dfrac{4}{3}$
A. $-2$
B. 2
C. $-\dfrac{4}{3}$
D. $\dfrac{4}{3}$
Đặt $u=\sqrt{3\cos x+1}\Rightarrow {{u}^{2}}=3\cos x+1\Rightarrow -\dfrac{2}{3}udu=\sin xdx$
Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{\pi }{2}\Rightarrow u=1 \\
& x=0\Rightarrow u=2 \\
\end{aligned} \right.$
Do đó $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin xf\left( \sqrt{3\cos x+1} \right)}{\sqrt{3\cos x+1}}}dx=\int\limits_{2}^{1}{\dfrac{-2uf\left( u \right)}{3u}}du=\dfrac{2}{3}\int\limits_{1}^{2}{f\left( u \right)du}=\dfrac{2}{3}\int\limits_{1}^{2}{f\left( x \right)dx}=\dfrac{4}{3}$
Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{\pi }{2}\Rightarrow u=1 \\
& x=0\Rightarrow u=2 \\
\end{aligned} \right.$
Do đó $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin xf\left( \sqrt{3\cos x+1} \right)}{\sqrt{3\cos x+1}}}dx=\int\limits_{2}^{1}{\dfrac{-2uf\left( u \right)}{3u}}du=\dfrac{2}{3}\int\limits_{1}^{2}{f\left( u \right)du}=\dfrac{2}{3}\int\limits_{1}^{2}{f\left( x \right)dx}=\dfrac{4}{3}$
Đáp án D.