Using the equation \[ \tan^{-1}x = \int_0^x \frac{dt}{1+t^2} \] show that, if \(x>0\), \(\tan^{-1}x\) lies between \(x-\displaystyle\frac{x^3}{3}\) and \(x-\displaystyle\frac{x^3}{3}+\frac{x^5}{5}\). Use this result to evaluate \(\tan^{-1}\frac{1}{11}\) correct to five places of decimals.