## University Calculus: Early Transcendentals (3rd Edition)

$\frac{1}{2}sin 2x+C$
Let I= $\int cos 2xdx$ Plug in: 2x=u $⇒dx= \frac{du}{2}$ Then: $I=\int cos u \frac{du}{2}$ $= \frac{1}{2}\int cos u du$ $= \frac{1}{2}\times sin u +C$ $= \frac{1}{2}sin 2x +C$