Definite integral via u-substitution

Let $u = \cos(2x)$, so $du = -2\sin(2x)\,dx$, which gives $\sin(2x)\,dx = -\tfrac{1}{2}du$. When $x=0$, $u=\cos(0)=1$; when $x=\pi/4$, $u=\cos(\pi/2)=0$. Substituting transforms the integral into $\int_1^0 e^{u}\left(-\tfrac12\right)du = \tfrac12\int_0^1 e^u\,du,$ where the sign flip comes from swapping the limits of integration.

Now evaluate $\tfrac12\int_0^1 e^u\,du = \tfrac12\left[e^u\right]_0^1 = \tfrac12(e^1 - e^0) = \tfrac12(e-1)$. This is the value of the original definite integral.