Evaluate $\int \sin 2x dx$

1. Analyze the question

You are asked to find the indefinite integral of the trigonometric function $\sin 2x$ with respect to $x$.

2. Explain the concept

To solve this, we use the rule for integrating trigonometric functions of the form $\sin(ax)$, where $a$ is a constant. The general rule is:

where $k$ is the constant of integration.

This rule is derived from the reverse of the chain rule in differentiation. Since the derivative of $\cos(ax)$ is $-a \sin(ax)$, the integral of $\sin(ax)$ must involve $-\frac{1}{a} \cos(ax)$.

3. Show the solution

• Step 1: Identify the constant $a$ in the expression $\sin 2x$. Here, $a = 2$.
• Step 2: Apply the integration formula: $$\int \sin 2x dx$$
• Step 3: Integrate $\sin(2x)$. The integral of $\sin$ is $-\cos$. Because the argument is $2x$, we must divide by the derivative of $2x$ (which is $2$): $$= -\frac{1}{2} \cos(2x) + k$$

4. Identify the answer

Comparing our result to the given options: A. $\cos 2x + k$ B. $\frac{1}{2}\cos 2x + k$ C. $-\frac{1}{2}\cos 2x + k$ D. $-\cos 2x + k$

The correct option is C.

5. Tip

Memory Trick: Remember the "C-Negative" rule for trigonometry integration. Whenever your integral results in a function starting with "co-" (like cosine), the result is negative.

• Integral of $\sin x = -\cos x$
• Integral of $\cos x = \sin x$