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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn $\int\limits_{0}^{\dfrac{\pi }{3}}{\tan x.f\left( {{\cos }^{2}}x \right)\text{d}x}=\int\limits_{1}^{8}{\dfrac{f\left( \sqrt[3]{x} \right)}{x}\text{d}x}=6$. Tính $\int\limits_{\dfrac{1}{2}}^{\sqrt{2}}{\dfrac{f\left( {{x}^{2}} \right)}{x}\text{d}x}$
A. $4$.
B. $6$.
C. $7$.
D. $10$.
A. $4$.
B. $6$.
C. $7$.
D. $10$.
Xét ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{3}}{\tan x.f\left( {{\cos }^{2}}x \right)\text{d}x}=6$.
Đặt $t={{\cos }^{2}}x\Rightarrow \text{d}t=-2\sin x.\cos x\text{d}x$. Đổi cận: $x=0\Rightarrow t=1;x=\dfrac{\pi }{3}\Rightarrow t=\dfrac{1}{4}$.
Khi đó: ${{I}_{1}}=-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{-2\sin x.\cos x}{{{\cos }^{2}}x}f\left( {{\cos }^{2}}x \right)\text{d}x}=-\dfrac{1}{2}\int\limits_{1}^{\dfrac{1}{4}}{\dfrac{f\left( t \right)}{t}\text{d}t}=\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( t \right)}{2t}\text{d}t}=6\Rightarrow \int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( x \right)}{2x}\text{d}x}=6$.
Xét ${{I}_{2}}=\int\limits_{1}^{8}{\dfrac{f\left( \sqrt[3]{x} \right)}{x}\text{d}x}=6$.
Đặt $t=\sqrt[3]{x}\Rightarrow {{t}^{3}}=x\Rightarrow 3{{t}^{2}}\text{d}t=\text{d}x$.
Khi $\left\{ \begin{aligned}
& x=1\Rightarrow t=1 \\
& x=8\Rightarrow t=2 \\
\end{aligned} \right. $. Ta có $ {{I}_{2}}=\int\limits_{1}^{2}{\dfrac{3{{t}^{2}}f\left( t \right)}{{{t}^{3}}}\text{d}t}=6\Rightarrow \int\limits_{1}^{2}{\dfrac{f\left( x \right)}{2x}\text{d}x}=1$.
Xét tích phân $I=\int\limits_{\dfrac{1}{2}}^{\sqrt{2}}{\dfrac{f\left( {{x}^{2}} \right)}{x}\text{d}x}~$.
Đặt $t={{x}^{2}}\Rightarrow \text{d}t=2x~\text{d}x$. Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{1}{2}\Rightarrow t=\dfrac{1}{4} \\
& x=\sqrt{2}\Rightarrow t=2 \\
\end{aligned} \right.$.
Ta có $I=\int\limits_{\dfrac{1}{2}}^{\sqrt{2}}{\dfrac{2xf\left( {{x}^{2}} \right)}{2{{x}^{2}}}\text{d}x}~=\int\limits_{\dfrac{1}{4}}^{2}{\dfrac{f\left( t \right)}{2t}\text{d}t}~=\int\limits_{\dfrac{1}{4}}^{2}{\dfrac{f\left( x \right)}{2x}\text{d}x}=\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( x \right)}{2x}\text{d}x}+\int\limits_{1}^{2}{\dfrac{f\left( x \right)}{2x}\text{d}x}=6+1=7$.
Đặt $t={{\cos }^{2}}x\Rightarrow \text{d}t=-2\sin x.\cos x\text{d}x$. Đổi cận: $x=0\Rightarrow t=1;x=\dfrac{\pi }{3}\Rightarrow t=\dfrac{1}{4}$.
Khi đó: ${{I}_{1}}=-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{-2\sin x.\cos x}{{{\cos }^{2}}x}f\left( {{\cos }^{2}}x \right)\text{d}x}=-\dfrac{1}{2}\int\limits_{1}^{\dfrac{1}{4}}{\dfrac{f\left( t \right)}{t}\text{d}t}=\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( t \right)}{2t}\text{d}t}=6\Rightarrow \int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( x \right)}{2x}\text{d}x}=6$.
Xét ${{I}_{2}}=\int\limits_{1}^{8}{\dfrac{f\left( \sqrt[3]{x} \right)}{x}\text{d}x}=6$.
Đặt $t=\sqrt[3]{x}\Rightarrow {{t}^{3}}=x\Rightarrow 3{{t}^{2}}\text{d}t=\text{d}x$.
Khi $\left\{ \begin{aligned}
& x=1\Rightarrow t=1 \\
& x=8\Rightarrow t=2 \\
\end{aligned} \right. $. Ta có $ {{I}_{2}}=\int\limits_{1}^{2}{\dfrac{3{{t}^{2}}f\left( t \right)}{{{t}^{3}}}\text{d}t}=6\Rightarrow \int\limits_{1}^{2}{\dfrac{f\left( x \right)}{2x}\text{d}x}=1$.
Xét tích phân $I=\int\limits_{\dfrac{1}{2}}^{\sqrt{2}}{\dfrac{f\left( {{x}^{2}} \right)}{x}\text{d}x}~$.
Đặt $t={{x}^{2}}\Rightarrow \text{d}t=2x~\text{d}x$. Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{1}{2}\Rightarrow t=\dfrac{1}{4} \\
& x=\sqrt{2}\Rightarrow t=2 \\
\end{aligned} \right.$.
Ta có $I=\int\limits_{\dfrac{1}{2}}^{\sqrt{2}}{\dfrac{2xf\left( {{x}^{2}} \right)}{2{{x}^{2}}}\text{d}x}~=\int\limits_{\dfrac{1}{4}}^{2}{\dfrac{f\left( t \right)}{2t}\text{d}t}~=\int\limits_{\dfrac{1}{4}}^{2}{\dfrac{f\left( x \right)}{2x}\text{d}x}=\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( x \right)}{2x}\text{d}x}+\int\limits_{1}^{2}{\dfrac{f\left( x \right)}{2x}\text{d}x}=6+1=7$.
Đáp án C.