It looks like the equation is
[tex]\dfrac{du}{dt} = 2t + \dfrac12 \sec^2(t) u[/tex]
with initial value [tex]u(0) = \frac\pi2[/tex].
Rearrange the equation to
[tex]\dfrac{du}{dt} - \dfrac12 \sec^2(t) u = 2t[/tex]
Multiply both sides by the integrating factor
[tex]\displaystyle \mu = \exp\left( \int -\frac12 \sec^2(t)\,dt\right) = \exp\left(-\frac12\tan(t)\right)[/tex]
and solve for [tex]u[/tex].
[tex]\implies e^{-\frac12 \tan(t)} \dfrac{du}{dt} - \dfrac12 \sec^2(t) e^{-\frac12 \tan(t)} u = 2t e^{-\frac12 \tan(t)}[/tex]
[tex]\implies \dfrac{d}{dt}\left[e^{-\frac12 \tan(t)} u\right] = 2t e^{-\frac12 \tan(t)}[/tex]
By the fundamental theorem of calculus, integrating both sides yields
[tex]e^{-\frac12\tan(t)} u = e^{-\frac12\tan(t)} u \bigg|_{t=0} + \displaystyle \int_{\xi=0}^{\xi=t} 2\xi e^{-\frac12 \tan(\xi)} \, d\xi[/tex]
[tex]\implies e^{-\frac12\tan(t)} u = 1\times\dfrac\pi2 + \displaystyle \int_{0}^{t} 2\xi e^{-\frac12 \tan(\xi)} \, d\xi[/tex]
[tex]\implies \boxed{\displaystyle u = \frac\pi2 e^{\frac12\tan(t)} + 2e^{\frac12\tan(t)} \int_0^t \xi e^{-\frac12 \tan(\xi)} \, d\xi}[/tex]
