If $x^{2}f\left(x\right)+f(1/x)=0$ for all $x\in \mathbf{R}-\left\{0\right\}$ then

${\int }_{\cos \theta }^{\sec \theta } f\left(x\right)dx=$

$\begin{array}{l}I={\int }_{\cos \theta }^{\sec \theta } f\left(x\right)dx=-{\int }_{\cos \theta }^{\sec \theta } \frac{1}{x^2}f\left(\frac{1}{x}\right)dx\left(\mathrm{Put}\frac{1}{x}=t\right)\\ ={\int }_{\sec \theta }^{\cos \theta } f\left(t\right)dt=-{\int }_{\cos \theta }^{\sec \theta } f\left(t\right)dt=-I\end{array}$

Hence $2I=0\Rightarrow I=0$