\( \left. \begin{array} { l } { \cos 2 x = 2 \cos ^ { 2 } x - 1 } \\ { \int \cos ^ { 2 } x } \end{array} \right. \)

Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int \cos^{2}\left(x\right) dx\) - step1: Evaluate the integral: \(\frac{x}{2}+\frac{\sin\left(2x\right)}{4}\) - step2: Simplify the expression: \(\frac{x}{2}+\frac{1}{4}\sin\left(2x\right)\) - step3: Add the constant of integral C: \(\frac{x}{2}+\frac{1}{4}\sin\left(2x\right) + C, C \in \mathbb{R}\) Let's break down the problem step by step. ### Step 1: Solve the equation \( \cos 2x = 2 \cos^2 x - 1 \) This equation is a trigonometric identity that relates the cosine of a double angle to the square of the cosine of the angle. 1. **Recognize the identity**: The equation \( \cos 2x = 2 \cos^2 x - 1 \) is a known trigonometric identity. It states that the cosine of double an angle \( 2x \) can be expressed in terms of the cosine of the angle \( x \). 2. **No further solving needed**: Since this is an identity, it holds true for all \( x \) where \( \cos x \) is defined. ### Step 2: Evaluate the integral \( \int \cos^2 x \, dx \) The integral of \( \cos^2 x \) can be solved using the power-reduction formula: \[ \cos^2 x = \frac{1 + \cos(2x)}{2} \] Now, we can rewrite the integral: \[ \int \cos^2 x \, dx = \int \frac{1 + \cos(2x)}{2} \, dx \] This can be split into two separate integrals: \[ = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \cos(2x) \, dx \] Calculating these integrals: 1. The integral of \( 1 \) is \( x \). 2. The integral of \( \cos(2x) \) is \( \frac{1}{2} \sin(2x) \). Putting it all together: \[ \int \cos^2 x \, dx = \frac{1}{2} x + \frac{1}{4} \sin(2x) + C \] where \( C \) is the constant of integration. ### Final Result 1. The equation \( \cos 2x = 2 \cos^2 x - 1 \) is an identity valid for all \( x \). 2. The integral \( \int \cos^2 x \, dx \) evaluates to: \[ \frac{x}{2} + \frac{1}{4} \sin(2x) + C \] where \( C \in \mathbb{R} \).