Cho hàm số $\Large f\left( x \right)$ liên tục trên $\Large \mathbb{R}

Cho hàm số $\Large f\left( x \right)$ liên tục trên $\Large \mathbb{R}$ và thỏa mãn $\Large \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot x.f\left( {{\sin }^{2}}x \right)\text{d}x}=\int\limits_{1}^{16}{\dfrac{f\left( \sqrt{x} \right)}{x}\text{d}x}=1$. Tính tích phân $\Large \int\limits_{\dfrac{1}{8}}^{1}{\dfrac{f\left( 4x \right)}{x}\text{d}x}$.

• A. $\Large I=3$.
• B. $\Large I=\dfrac{3}{2}$.
• C. $\Large I=2$.
• D. $\Large I=\dfrac{5}{2}$.

Đặt $\Large {{I}_{1}}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot x.f\left( {{\sin }^{2}}x \right)\text{d}x}=1$, $\Large {{I}_{2}}=\int\limits_{1}^{16}{\dfrac{f\left( \sqrt{x} \right)}{x}\text{d}x}=1$.
Đặt $\Large t={{\sin }^{2}}x$ $\Large \Rightarrow \text{d}t=2\sin x.\cos x\text{d}x$$\Large=2{{\sin }^{2}}x.\cot x\text{d}x$$\Large=2t.\cot x\text{d}x$.
 
$\Large{{I}_{1}}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot x.f\left( {{\sin }^{2}}x \right)\text{d}x}$$=\int\limits_{\dfrac{1}{2}}^{1}{f\left( t \right).\dfrac{1}{2t}\text{d}t}$ $\Large =\dfrac{1}{2}\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{f\left( t \right)}{t}\text{d}t}$$\Large=\dfrac{1}{2}\int\limits_{\dfrac{1}{8}}^{\dfrac{1}{4}}{\dfrac{f\left( 4x \right)}{4x}\text{d}\left( 4x \right)}$$\Large=\dfrac{1}{2}\int\limits_{\dfrac{1}{8}}^{\dfrac{1}{4}}{\dfrac{f\left( 4x \right)}{x}\text{d}x}$.
Suy ra $\Large \int\limits_{\dfrac{1}{8}}^{\dfrac{1}{4}}{\dfrac{f\left( 4x \right)}{x}\text{d}x}=2{{I}_{1}}=2$
•Đặt $\Large t=\sqrt{x}$ $\Large \Rightarrow 2t\text{d}t=\text{d}x$.
 
$\Large {{I}_{2}}=\int\limits_{1}^{16}{\dfrac{f\left( \sqrt{x} \right)}{x}\text{d}x}$ $\Large =\int\limits_{1}^{4}{\dfrac{f\left( t \right)}{{{t}^{2}}}\text{2}t\text{d}t}$ $\Large =2\int\limits_{1}^{4}{\dfrac{f\left( t \right)}{t}\text{d}t}$ $\Large =2\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( 4x \right)}{4x}\text{d}\left( 4x \right)}$ $\Large =2\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( 4x \right)}{x}\text{d}x}$.
Suy ra $\Large \int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( 4x \right)}{x}\text{d}x}=\dfrac{1}{2}{{I}_{2}}=\dfrac{1}{2}$
Khi đó, ta có:
$\Large \int\limits_{\dfrac{1}{8}}^{1}{\dfrac{f\left( 4x \right)}{x}\text{d}x}=\int\limits_{\dfrac{1}{8}}^{\dfrac{1}{4}}{\dfrac{f\left( 4x \right)}{x}\text{d}x}+\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{f\left( 4x \right)}{x}\text{d}x}$$\Large=2+\dfrac{1}{2}=\dfrac{5}{2}$.