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Câu hỏi: Biết $\int\limits_{0}^{\dfrac{5}{2}}{\sqrt{\dfrac{5+x}{5-x}} \text{d}x}-\dfrac{5\pi }{6}=\dfrac{5\left( a-\sqrt{b} \right)}{2}$ với $a, b\in \mathbb{N}.$ Tính $T=a+2b.$
A. $T=8.$
B. $T=6.$
C. $T=7.$
D. $T=5.$
A. $T=8.$
B. $T=6.$
C. $T=7.$
D. $T=5.$
Đặt $x=5\cos 2t \Rightarrow dx=-10\sin 2t.$
Đổi cận $x=0 \Rightarrow t=\dfrac{\pi }{4} ; x=\dfrac{5}{2} \Rightarrow t=\dfrac{\pi }{6}.$
Do đó $\int\limits_{0}^{\dfrac{5}{2}}{\sqrt{\dfrac{5+x}{5-x}} dx} =-10\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}}{\sqrt{\dfrac{5\left( 1+\cos 2t \right)}{5\left( 1-\cos 2t \right)}}\sin 2t dt}=10\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\dfrac{\cos t}{\sin t}2\sin t\cos t }dt$
$=10\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\left( 1+\cos 2t \right)dt}=10\left. \left( t+\dfrac{1}{2}\sin 2t \right) \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}=10\left( \dfrac{\pi }{4}+\dfrac{1}{2}-\dfrac{\pi }{6}-\dfrac{\sqrt{3}}{4} \right)$
$=10\left( \dfrac{\pi }{12}-\dfrac{2-\sqrt{3}}{4} \right)=\dfrac{5\pi }{6}+\dfrac{5\left( 2-\sqrt{3} \right)}{2}.$
Suy ra $a=2, b=3.$ Vậy $T=2+2.3=8.$
Đổi cận $x=0 \Rightarrow t=\dfrac{\pi }{4} ; x=\dfrac{5}{2} \Rightarrow t=\dfrac{\pi }{6}.$
Do đó $\int\limits_{0}^{\dfrac{5}{2}}{\sqrt{\dfrac{5+x}{5-x}} dx} =-10\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}}{\sqrt{\dfrac{5\left( 1+\cos 2t \right)}{5\left( 1-\cos 2t \right)}}\sin 2t dt}=10\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\dfrac{\cos t}{\sin t}2\sin t\cos t }dt$
$=10\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\left( 1+\cos 2t \right)dt}=10\left. \left( t+\dfrac{1}{2}\sin 2t \right) \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}=10\left( \dfrac{\pi }{4}+\dfrac{1}{2}-\dfrac{\pi }{6}-\dfrac{\sqrt{3}}{4} \right)$
$=10\left( \dfrac{\pi }{12}-\dfrac{2-\sqrt{3}}{4} \right)=\dfrac{5\pi }{6}+\dfrac{5\left( 2-\sqrt{3} \right)}{2}.$
Suy ra $a=2, b=3.$ Vậy $T=2+2.3=8.$
Đáp án A.