What is the result of ∫ sin^2(x) dx?

What is the result of ∫ sin^2(x) dx?

• A. (1/2)(x + sin(2x)/2) + C
• B. (1/2)(x - (1/2)sin(2x)) + C
• C. sin(x) + C
• D. (1/3)(x^3) + C

Answer: (B)

To find the integral of sin²(x) with respect to x, we can use the identity involving the double angle formula for cosine. The identity states that sin²(x) can be rewritten as: \[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \] Now, substituting this back into the integral: \[ \int \sin^2(x) \, dx = \int \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int (1 - \cos(2x)) \, dx \] This integral can be separated into two parts: \[ \int (1 - \cos(2x)) \, dx = \int 1 \, dx - \int \cos(2x) \, dx \] Calculating each integral separately: 1. The integral of 1 with respect to x is simply x. 2. For the integral of cos(2x), we apply the substitution u = 2x, leading to du = 2dx, or dx = (1/2)du. Thus, \[ \int \cos(

To find the integral of sin²(x) with respect to x, we can use the identity involving the double angle formula for cosine. The identity states that sin²(x) can be rewritten as:

[

\sin^2(x) = \frac{1 - \cos(2x)}{2}

]

Now, substituting this back into the integral:

[

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

]

This integral can be separated into two parts:

[

\int (1 - \cos(2x)) , dx = \int 1 , dx - \int \cos(2x) , dx

]

Calculating each integral separately:

1. The integral of 1 with respect to x is simply x.

2. For the integral of cos(2x), we apply the substitution u = 2x, leading to du = 2dx, or dx = (1/2)du. Thus,

[

\int \cos(