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Câu hỏi: Biết $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{{{\cos }^{2}}x+\sin x\cos x+1}{{{\cos }^{4}}x+\sin x{{\cos }^{3}}x}\text{d}x}=a+b\ln 2+c\ln \left( 1+\sqrt{3} \right)$, với $a,b,c$ là các số hữu tỉ. Giá trị của $abc$ bằng
A. $0$.
B. $-2$.
C. $-4$.
D. $-6$.
Ta có: $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{{{\cos }^{2}}x+\sin x\cos x+1}{{{\cos }^{4}}x+\sin x{{\cos }^{3}}x}\text{d}x}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{\dfrac{1}{{{\cos }^{2}}x}+\dfrac{\tan x}{{{\cos }^{2}}x}+\dfrac{1}{{{\cos }^{4}}x}}{1+\tan x}\text{d}x}$
$=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{\left( 1+{{\tan }^{2}}x \right)+\tan x\left( 1+{{\tan }^{2}}x \right)+{{\left( 1+{{\tan }^{2}}x \right)}^{2}}}{1+\tan x}\text{d}x}$
$=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{1+\tan x+\left( 1+{{\tan }^{2}}x \right)}{1+\tan x}\left( 1+{{\tan }^{2}}x \right)\text{d}x}$ $=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\left( 1+\dfrac{1+{{\tan }^{2}}x}{1+\tan x} \right)\left( 1+{{\tan }^{2}}x \right)\text{d}x}$.
Đặt $t=1+\tan x$ ta được $\text{d}t=\left( 1+{{\tan }^{2}}x \right)\text{d}x$, đổi cận $x=\dfrac{\pi }{4}\Rightarrow t=2,\ x=\dfrac{\pi }{3}\Rightarrow t=1+\sqrt{3}$
Ta được
$\int\limits_{2}^{1+\sqrt{3}}{\left( 1+\dfrac{1+{{\left( t-1 \right)}^{2}}}{t} \right)\text{d}t}=\int\limits_{2}^{1+\sqrt{3}}{\left( t-1+\dfrac{2}{t} \right)\text{d}t}=\left. \left( \dfrac{{{t}^{2}}}{2}-t+2\ln t \right) \right|_{2}^{1+\sqrt{3}}=1-2\ln 2+2\ln \left( 1+\sqrt{3} \right)$
Từ đây ta suy ra $a+b\ln 2+c\ln \left( 1+\sqrt{3} \right)=1-2\ln 2+2\ln \left( 1+\sqrt{3} \right)$.
Do đó $a=1,b=-2,c=2$ suy ra $abc=-4.$
A. $0$.
B. $-2$.
C. $-4$.
D. $-6$.
Ta có: $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{{{\cos }^{2}}x+\sin x\cos x+1}{{{\cos }^{4}}x+\sin x{{\cos }^{3}}x}\text{d}x}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{\dfrac{1}{{{\cos }^{2}}x}+\dfrac{\tan x}{{{\cos }^{2}}x}+\dfrac{1}{{{\cos }^{4}}x}}{1+\tan x}\text{d}x}$
$=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{\left( 1+{{\tan }^{2}}x \right)+\tan x\left( 1+{{\tan }^{2}}x \right)+{{\left( 1+{{\tan }^{2}}x \right)}^{2}}}{1+\tan x}\text{d}x}$
$=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\dfrac{1+\tan x+\left( 1+{{\tan }^{2}}x \right)}{1+\tan x}\left( 1+{{\tan }^{2}}x \right)\text{d}x}$ $=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{\left( 1+\dfrac{1+{{\tan }^{2}}x}{1+\tan x} \right)\left( 1+{{\tan }^{2}}x \right)\text{d}x}$.
Đặt $t=1+\tan x$ ta được $\text{d}t=\left( 1+{{\tan }^{2}}x \right)\text{d}x$, đổi cận $x=\dfrac{\pi }{4}\Rightarrow t=2,\ x=\dfrac{\pi }{3}\Rightarrow t=1+\sqrt{3}$
Ta được
$\int\limits_{2}^{1+\sqrt{3}}{\left( 1+\dfrac{1+{{\left( t-1 \right)}^{2}}}{t} \right)\text{d}t}=\int\limits_{2}^{1+\sqrt{3}}{\left( t-1+\dfrac{2}{t} \right)\text{d}t}=\left. \left( \dfrac{{{t}^{2}}}{2}-t+2\ln t \right) \right|_{2}^{1+\sqrt{3}}=1-2\ln 2+2\ln \left( 1+\sqrt{3} \right)$
Từ đây ta suy ra $a+b\ln 2+c\ln \left( 1+\sqrt{3} \right)=1-2\ln 2+2\ln \left( 1+\sqrt{3} \right)$.
Do đó $a=1,b=-2,c=2$ suy ra $abc=-4.$
Đáp án C.