`int(1)/(cos2x+sin^(2)x)dx=`

Options

A. `sinx+c`
B. `cosx+c`
C. `tanx`
D. `cotx`

Answer

To solve the integral \( \int \frac{1}{\cos 2x + \sin^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the integral:
\[ \int \frac{1}{\cos 2x + \sin^2 x} \, dx \]
We know that \( \cos 2x = \cos^2 x - \sin^2 x \). Therefore, we can rewrite the integral as:
\[ \int \frac{1}{(\cos^2 x - \sin^2 x) + \sin^2 x} \, dx \]
This simplifies to:
\[ \int \frac{1}{\cos^2 x} \, dx \]

### Step 2: Simplify the integral
The expression \( \frac{1}{\cos^2 x} \) can be rewritten as:
\[ \sec^2 x \]
Thus, our integral now becomes:
\[ \int \sec^2 x \, dx \]

### Step 3: Integrate
The integral of \( \sec^2 x \) is a standard integral:
\[ \int \sec^2 x \, dx = \tan x + C \]
where \( C \) is the constant of integration.

### Final Answer
Putting it all together, we have:
\[ \int \frac{1}{\cos 2x + \sin^2 x} \, dx = \tan x + C \]