Find the integral of 2sin(x)cos(x) dx.

Find the integral of 2sin(x)cos(x) dx.

• A. sin^2(x) + C
• B. 2cos^2(x) + C
• C. sin(x)cos(x) + C
• D. -sin^2(x) + C

Answer: (A)

To find the integral of \(2\sin(x)\cos(x) \, dx\), we can use a well-known trigonometric identity. Specifically, the double angle identity states that: \[ \sin(2x) = 2\sin(x)\cos(x). \] Using this identity, we can rewrite the integral as: \[ \int 2\sin(x)\cos(x) \, dx = \int \sin(2x) \, dx. \] The integral of \(\sin(2x)\) can be solved using a substitution. If we let \(u = 2x\), then \(du = 2dx\) or \(dx = \frac{du}{2}\). Substituting into the integral gives: \[ \int \sin(2x) \, dx = \frac{1}{2} \int \sin(u) \, du. \] The integral of \(\sin(u)\) is \(-\cos(u) + C\), so we have: \[ \frac{1}{2}(-\cos(u)) + C = -\frac{1}{2}\cos(2x) + C. \

To find the integral of (2\sin(x)\cos(x) , dx), we can use a well-known trigonometric identity. Specifically, the double angle identity states that:

[

\sin(2x) = 2\sin(x)\cos(x).

]

Using this identity, we can rewrite the integral as:

[

\int 2\sin(x)\cos(x) , dx = \int \sin(2x) , dx.

]

The integral of (\sin(2x)) can be solved using a substitution. If we let (u = 2x), then (du = 2dx) or (dx = \frac{du}{2}). Substituting into the integral gives:

[

\int \sin(2x) , dx = \frac{1}{2} \int \sin(u) , du.

]

The integral of (\sin(u)) is (-\cos(u) + C), so we have:

[

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

\