Answer
$$\int_0^\frac{\pi}{4}\cos{2x}\hspace{0.5mm}dx=\frac{1}{2}$$
Work Step by Step
This integral is a very simple application of u-substitution, and remembering that the antiderivative of $\cos{x}$ is $\sin{x}.$ By choosing $u=2x$ we then see that $du=2dx$ and so then $\frac{1}{2}du=dx$. Substituting these back into our original integral we get $$\int \frac{1}{2}\cos{u}\hspace{0.5mm}du.$$
Remembering that we can take constants out of integrals we get $$ \frac{1}{2}\int\cos{u}\hspace{0.5mm}du=\frac{1}{2}\left(\sin{u}+C\right).$$
Now we can resubstitute $u=2x$. Therefore we get, $$\int \cos{2x}\hspace{0.5mm}dx=\frac{1}{2}\left(\sin{2x}+C\right)$$Now all we have left to do is plug in our integration limits.
$$\frac{1}{2}\left(\sin{2x}+C\right)\bigg\vert_0^{\frac{\pi}{4}}=\frac{1}{2}\left(\sin{(2\times\frac{\pi}{4})}+C\right)-\frac{1}{2}\left(\sin{0}+C\right).$$ We can remember that the constant of integration $C$ can be ignored. Applying this and remembering that $\sin{0}=0$. We can simplify our expression $$\frac{1}{2}\left(\sin{(2\times\frac{\pi}{4})}+C\right)-\frac{1}{2}\left(\sin{0}+C\right)=\sin{(2\times\frac{\pi}{4})}$$
$$=\sin{\frac{\pi}{2}}=1$$
