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Câu hỏi: Cho $\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sqrt{2+3\tan x}}{1+\cos 2x}dx}=a\sqrt{5}+b\sqrt{2}$, với $a,b\in \mathbb{R}$. Tính giá trị biểu thức A = a + b.
A. $\dfrac{1}{3}.$
B. $\dfrac{7}{12}.$
C. $\dfrac{2}{3}.$
D. $\dfrac{4}{3}.$
A. $\dfrac{1}{3}.$
B. $\dfrac{7}{12}.$
C. $\dfrac{2}{3}.$
D. $\dfrac{4}{3}.$
Ta có $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sqrt{2+3\tan x}}{1+\cos 2x}dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sqrt{2+3\tan x}}{2{{\cos }^{2}}x}dx}$
Đặt $u=\sqrt{2+3\tan x}\Rightarrow {{u}^{2}}=2+3\tan x\Rightarrow 2udu=\dfrac{3}{{{\cos }^{2}}x}dx$
Đổi cận: $x=0\Rightarrow u=\sqrt{2}.$
$x=\dfrac{\pi }{4}\Rightarrow u=\sqrt{5}.$
Khi đó $I=\dfrac{1}{3}\int\limits_{\sqrt{2}}^{\sqrt{5}}{{{u}^{2}}du=}\left. \dfrac{1}{9}{{u}^{3}} \right|_{\sqrt{2}}^{\sqrt{5}}=\dfrac{5\sqrt{5}}{9}-\dfrac{2\sqrt{2}}{9}$. Do đó $a=\dfrac{5}{9},b=-\dfrac{2}{9}\Rightarrow a+b=\dfrac{1}{3}.$
Đặt $u=\sqrt{2+3\tan x}\Rightarrow {{u}^{2}}=2+3\tan x\Rightarrow 2udu=\dfrac{3}{{{\cos }^{2}}x}dx$
Đổi cận: $x=0\Rightarrow u=\sqrt{2}.$
$x=\dfrac{\pi }{4}\Rightarrow u=\sqrt{5}.$
Khi đó $I=\dfrac{1}{3}\int\limits_{\sqrt{2}}^{\sqrt{5}}{{{u}^{2}}du=}\left. \dfrac{1}{9}{{u}^{3}} \right|_{\sqrt{2}}^{\sqrt{5}}=\dfrac{5\sqrt{5}}{9}-\dfrac{2\sqrt{2}}{9}$. Do đó $a=\dfrac{5}{9},b=-\dfrac{2}{9}\Rightarrow a+b=\dfrac{1}{3}.$
Đáp án A.