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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 $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sqrt{2+3\tan x}}{1+\cos 2x}\text{d}x}$ $=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sqrt{2+3\tan x}}{2{{\cos }^{2}}x}\text{d}x}$
Đặt $u=\sqrt{2+3\tan x}$ $\Rightarrow {{u}^{2}}=2+3\tan x$ $\Rightarrow 2u\text{d}u=\dfrac{3}{{{\cos }^{2}}x}\text{d}x$
Đổi cận $x=0\Rightarrow u=\sqrt{2}$
$x=\dfrac{\pi }{4}\Rightarrow u=\sqrt{5}$.
Khi đó $I=\dfrac{1}{3}\int\limits_{\sqrt{2}}^{\sqrt{5}}{{{u}^{2}}\text{d}u}=\dfrac{1}{9}\left. {{u}^{3}} \right|_{\sqrt{2}}^{\sqrt{5}}$ $=\dfrac{5\sqrt{5}}{9}-\dfrac{2\sqrt{2}}{9}$.
Do đó $a=\dfrac{5}{9}$, $b=-\dfrac{2}{9}$ $a=\dfrac{5}{9}, b=-\dfrac{2}{9}\Rightarrow a+b=\dfrac{1}{3}$.
Đặt $u=\sqrt{2+3\tan x}$ $\Rightarrow {{u}^{2}}=2+3\tan x$ $\Rightarrow 2u\text{d}u=\dfrac{3}{{{\cos }^{2}}x}\text{d}x$
Đổi cận $x=0\Rightarrow u=\sqrt{2}$
$x=\dfrac{\pi }{4}\Rightarrow u=\sqrt{5}$.
Khi đó $I=\dfrac{1}{3}\int\limits_{\sqrt{2}}^{\sqrt{5}}{{{u}^{2}}\text{d}u}=\dfrac{1}{9}\left. {{u}^{3}} \right|_{\sqrt{2}}^{\sqrt{5}}$ $=\dfrac{5\sqrt{5}}{9}-\dfrac{2\sqrt{2}}{9}$.
Do đó $a=\dfrac{5}{9}$, $b=-\dfrac{2}{9}$ $a=\dfrac{5}{9}, b=-\dfrac{2}{9}\Rightarrow a+b=\dfrac{1}{3}$.
Đáp án A.