[tex]\displaystyle
\frac{du}{dt} = \frac{2t + \sec^2 t}{2u}\ \Rightarrow\ 2u\, du = \left( 2t + \sec^2 t \right) dt\ \Rightarrow\\ \\ \textstyle\int 2u\, du = \int \left( 2t + \sec^2 t \right) dt\ \Rightarrow\ u^2 = t^2 + \tan t + C, \\ \\
\text{where } \big[u(0\big]^2 = 0^2 + \tan 0 + C\ \Rightarrow\ C = (-5)^2 = 25. \\ \\
\text{Therefore, } u^2 = t^2 + \tan t + 25,\text{ so } u = \pm \sqrt{t^2 + \tan t + 25}. \\ \\
\text{Since } u(0) = -5\text{, we must have }\ u = - \sqrt{t^2 + \tan t + 25}.[/tex]
