Xét \({{I}_{1}}=\int\limits_{0}^{\frac{\pi }{3}}{\tan \,x.f\left( {{\cos }^{2}}x \right)\text{d}x}=6\).

Đặt \(t={{\cos }^{2}}x\Rightarrow \text{d}t=-2\sin x.\cos x\text{d}x\). Đổi cận: \(x=0\Rightarrow t=1;x=\frac{\pi }{3}\Rightarrow t=\frac{1}{4}\).

Khi đó: \({{I}_{1}}=-\frac{1}{2}\int\limits_{0}^{\frac{\pi }{3}}{\frac{-2\sin x.\cos x}{{{\cos }^{2}}x}f\left( {{\cos }^{2}}x \right)\text{d}x}=-\frac{1}{2}\int\limits_{1}^{\frac{1}{4}}{\frac{f\left( t \right)}{t}\text{d}t}=\int\limits_{\frac{1}{4}}^{1}{\frac{f\left( t \right)}{2t}\text{d}t}=6\Rightarrow \int\limits_{\frac{1}{4}}^{1}{\frac{f\left( x \right)}{2x}\text{d}x}=6\)

Xét \({{I}_{2}}=\int\limits_{1}^{8}{\frac{f\left( \sqrt[3]{x} \right)}{x}\text{d}x}=6\).

Đặt \(t=\sqrt[3]{x}\Rightarrow {{t}^{3}}=x\Rightarrow 3{{t}^{2}}\text{d}t=\text{d}x\).

Khi \(\left\{ \begin{align} & x=1\Rightarrow t=1 \\ & x=8\Rightarrow t=2 \\ \end{align} \right.\). Ta có \({{I}_{2}}=\int\limits_{1}^{2}{\frac{3{{t}^{2}}f\left( t \right)}{{{t}^{3}}}\text{d}t}=6\Rightarrow \int\limits_{1}^{2}{\frac{f\left( x \right)}{2x}\text{d}x}=1\).

Xét tích phân \(I=\int\limits_{\frac{1}{2}}^{\sqrt{2}}{\frac{f\left( {{x}^{2}} \right)}{x}\text{d}x}~\).

Đặt \(t={{x}^{2}}\Rightarrow \text{d}t=2x~\text{d}x\). Đổi cận \(\left\{ \begin{align} & x=\frac{1}{2}\Rightarrow t=\frac{1}{4} \\ & x=\sqrt{2}\Rightarrow t=2 \\ \end{align} \right.\).

Ta có \(I=\int\limits_{\frac{1}{2}}^{\sqrt{2}}{\frac{2xf\left( {{x}^{2}} \right)}{2{{x}^{2}}}\text{d}x}~=\int\limits_{\frac{1}{4}}^{2}{\frac{f\left( t \right)}{2t}\text{d}t}~=\int\limits_{\frac{1}{4}}^{2}{\frac{f\left( x \right)}{2x}\text{d}x}=\int\limits_{\frac{1}{4}}^{1}{\frac{f\left( x \right)}{2x}\text{d}x}+\int\limits_{1}^{2}{\frac{f\left( x \right)}{2x}\text{d}x}=6+1=7\).