What is the result of ∫ (1 + cos(2x)) dx?

What is the result of ∫ (1 + cos(2x)) dx?

• A. x + (1/2)sin(2x) + C
• B. 2x + sin(2x) + C
• C. x + sin(2x) + C
• D. sin(2x) + C

Answer: (A)

To determine the integral of the function \( \int (1 + \cos(2x)) \, dx \), we can break the integrand into two parts and evaluate each separately. First, we handle the constant term \( 1 \). The integral of \( 1 \) with respect to \( x \) is simply \( x \): \[ \int 1 \, dx = x. \] Next, we need to integrate \( \cos(2x) \). To do this, we can use the substitution method. The integral of \( \cos(kx) \) is given by \( \frac{1}{k} \sin(kx) \). Here, \( k \) is \( 2 \): \[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x). \] Now, combining both results, we find: \[ \int (1 + \cos(2x)) \, dx = \int 1 \, dx + \int \cos(2x) \, dx = x + \frac{1}{2} \sin(2x). \] Finally, we add the constant

To determine the integral of the function ( \int (1 + \cos(2x)) , dx ), we can break the integrand into two parts and evaluate each separately.

First, we handle the constant term ( 1 ). The integral of ( 1 ) with respect to ( x ) is simply ( x ):

[

\int 1 , dx = x.

]

Next, we need to integrate ( \cos(2x) ). To do this, we can use the substitution method. The integral of ( \cos(kx) ) is given by ( \frac{1}{k} \sin(kx) ). Here, ( k ) is ( 2 ):

[

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

]

Now, combining both results, we find:

[

\int (1 + \cos(2x)) , dx = \int 1 , dx + \int \cos(2x) , dx = x + \frac{1}{2} \sin(2x).

]

Finally, we add the constant