Question

1. if \\( \frac{d y}{d x}=\sin (2 x) \\), which of the following is a valid choice for \\( y \\)?

a. \\( \cos (2 x)+c \\)
b. \\( \frac{1}{2} \cos (2 x)+c \\)
c. \\( -\frac{1}{2} \cos (2 x)+c \\)
d. \\( \frac{1}{2} \sin (2 x)+c \\)
e. \\( -\frac{1}{2} \sin (2 x)+c \\)
evaluate the limits in

Explanation:

Step1: Integrate \(\frac{dy}{dx}=\sin(2x)\)

We know that \(\int\sin(ax)dx =-\frac{1}{a}\cos(ax)+C\) (where \(a = 2\) in our case).
Using the formula \(\int\sin(2x)dx\), let \(u = 2x\), then \(du=2dx\) and \(dx=\frac{1}{2}du\).
So \(\int\sin(2x)dx=\frac{1}{2}\int\sin(u)du\).
Since \(\int\sin(u)du=-\cos(u)+C\), substituting back \(u = 2x\) gives \(\frac{1}{2}(-\cos(2x))+C=-\frac{1}{2}\cos(2x)+C\).

Answer:

C. \(-\frac{1}{2}\cos(2x)+C\)