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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 $a,b\in \mathbb{R}$. Giá trị biểu thức $A=a+b$ là
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 $I=\dfrac{1}{3}\int\limits_{\sqrt{2}}^{\sqrt{5}}{{{u}^{2}}du}=\dfrac{1}{9}{{u}^{3}}\left| \begin{aligned}
& ^{\sqrt{5}} \\
& _{\sqrt{2}} \\
\end{aligned} \right.=\dfrac{5\sqrt{5}}{9}-\dfrac{2\sqrt{2}}{9}$
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 $I=\dfrac{1}{3}\int\limits_{\sqrt{2}}^{\sqrt{5}}{{{u}^{2}}du}=\dfrac{1}{9}{{u}^{3}}\left| \begin{aligned}
& ^{\sqrt{5}} \\
& _{\sqrt{2}} \\
\end{aligned} \right.=\dfrac{5\sqrt{5}}{9}-\dfrac{2\sqrt{2}}{9}$
Do đó $a=\dfrac{5}{9},b=-\dfrac{2}{9}\Rightarrow a+b=\dfrac{1}{3}$.
Đáp án A.