Evaluate the following integrals:
`int ( cos 2x+2sin^(2)x)/(cos^(2)x) dx`

Answer

To evaluate the integral

\[ \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} \, dx, \]

we can start by simplifying the integrand.

### Step 1: Rewrite \(\cos 2x\)
We know that

\[ \cos 2x = \cos^2 x - \sin^2 x. \]

However, we can also use the identity

\[ \cos 2x = 1 - 2\sin^2 x. \]

Using this identity, we can rewrite the integrand:

\[ \cos 2x + 2\sin^2 x = (1 - 2\sin^2 x) + 2\sin^2 x = 1. \]

### Step 2: Substitute back into the integral
Now substituting this back into the integral, we have:

\[ \int \frac{1}{\cos^2 x} \, dx. \]

### Step 3: Recognize the integrand
We know that

\[ \frac{1}{\cos^2 x} = \sec^2 x. \]

Thus, our integral simplifies to:

\[ \int \sec^2 x \, dx. \]

### Step 4: Integrate
The integral of \(\sec^2 x\) is a standard result:

\[ \int \sec^2 x \, dx = \tan x + C, \]

where \(C\) is the constant of integration.

### Final Answer
Therefore, the final result of the integral is:

\[ \tan x + C. \]

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