Answer
$$w = {\tan ^{ - 1}}\left( {{t^2} + 1} \right)$$
Work Step by Step
$$\eqalign{
& w'\left( t \right) = 2t{\cos ^2}w \cr
& \frac{{dw}}{{dt}} = 2t{\cos ^2}w \cr
& {\text{separate}} \cr
& \frac{{dw}}{{{{\cos }^2}w}} = 2tdt \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{dw}}{{{{\cos }^2}w}}} = \int {2tdt} \cr
& \int {{{\sec }^2}w} = \int {2tdt} \cr
& \tan w = {t^2} + C \cr
& {\text{the initial condition }}w\left( 0 \right) = \frac{\pi }{4}{\text{ implies that}} \cr
& \tan \frac{\pi }{4} = {\left( 0 \right)^2} + C \cr
& C = 1 \cr
& \tan w = {t^2} + C \cr
& \tan w = {t^2} + 1 \cr
& {\text{solve for }}w \cr
& {\tan ^{ - 1}}\left( {\tan w} \right) = {\tan ^{ - 1}}\left( {{t^2} + 1} \right) \cr
& w = {\tan ^{ - 1}}\left( {{t^2} + 1} \right) \cr} $$
