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

Đặt ${{I}_{1}}=\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cot x.f\left( {{\sin }^{2}}x \right)\text{d}x}=1$, ${{I}_{2}}=\int\limits_{1}^{16}{\frac{f\left( \sqrt{x} \right)}{x}\text{d}x}=1$.

Đặt $t={{\sin }^{2}}x \Rightarrow \text{d}t=2\sin x.\cos x\text{d}x =2{{\sin }^{2}}x.\cot x\text{d}x =2t.\cot x\text{d}x$.

${{I}_{1}}=\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cot x.f\left( {{\sin }^{2}}x \right)\text{d}x}=\int\limits_{\frac{1}{2}}^{1}{f\left( t \right).\frac{1}{2t}\text{d}t} =\frac{1}{2}\int\limits_{\frac{1}{2}}^{1}{\frac{f\left( t \right)}{t}\text{d}t} =\frac{1}{2}\int\limits_{\frac{1}{8}}^{\frac{1}{4}}{\frac{f\left( 4x \right)}{4x}\text{d}\left( 4x \right)} =\frac{1}{2}\int\limits_{\frac{1}{8}}^{\frac{1}{4}}{\frac{f\left( 4x \right)}{x}\text{d}x}$.

Suy ra $\int\limits_{\frac{1}{8}}^{\frac{1}{4}}{\frac{f\left( 4x \right)}{x}\text{d}x}=2{{I}_{1}}=2$

Đặt $t=\sqrt{x} \Rightarrow 2t\text{d}t=\text{d}x$.

${{I}_{2}}=\int\limits_{1}^{16}{\frac{f\left( \sqrt{x} \right)}{x}\text{d}x} =\int\limits_{1}^{4}{\frac{f\left( t \right)}{{{t}^{2}}}\text{2}t\text{d}t} =2\int\limits_{1}^{4}{\frac{f\left( t \right)}{t}\text{d}t} =2\int\limits_{\frac{1}{4}}^{1}{\frac{f\left( 4x \right)}{4x}\text{d}\left( 4x \right)} =2\int\limits_{\frac{1}{4}}^{1}{\frac{f\left( 4x \right)}{x}\text{d}x}$.

Suy ra $\int\limits_{\frac{1}{4}}^{1}{\frac{f\left( 4x \right)}{x}\text{d}x}=\frac{1}{2}{{I}_{2}}=\frac{1}{2}$

Khi đó, ta có: $\int\limits_{\frac{1}{8}}^{1}{\frac{f\left( 4x \right)}{x}\text{d}x}=\int\limits_{\frac{1}{8}}^{\frac{1}{4}}{\frac{f\left( 4x \right)}{x}\text{d}x}+\int\limits_{\frac{1}{4}}^{1}{\frac{f\left( 4x \right)}{x}\text{d}x} =2+\frac{1}{2}=\frac{5}{2}$.