Question

Find an antiderivative of \[\cos 2x.\]

Options

• \[2\sin 2x+C\]

• \[\sin 2x+C\]

• \[\frac{1}{2}\sin 2x+C\]

• \[-\frac{1}{2}\cos 2x+C\]

Answer 1

Since \[\frac{d}{dx}(\sin 2x)=2\cos 2x,\] a factor of \[\frac{1}{2}\] is required. Therefore, \[\int\cos 2x\,dx=\frac{1}{2}\sin 2x+C.\]