Evaluate the following integrals:
`int(dx)/((2sinx+cosx+3))`

Explanation

To evaluate the integral

\[ I = \int \frac{dx}{2 \sin x + \cos x + 3}, \]

we will use the substitution \( t = \tan\left(\frac{x}{2}\right) \). This substitution will help us express \( \sin x \) and \( \cos x \) in terms of \( t \).

### Step 1: Use the Half-Angle Formulas
Using the half-angle formulas, we have:

\[ \sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}. \]

### Step 2: Substitute \( \sin x \) and \( \cos x \) into the Integral
Substituting these into the integral gives:

\[ I = \int \frac{dx}{2\left(\frac{2t}{1+t^2}\right) + \left(\frac{1-t^2}{1+t^2}\right) + 3}. \]

### Step 3: Simplify the Denominator
The denominator simplifies as follows:

\[ 2\left(\frac{2t}{1+t^2}\right) + \left(\frac{1-t^2}{1+t^2}\right) + 3 = \frac{4t + 1 - t^2 + 3(1+t^2)}{1+t^2} = \frac{4t + 1 - t^2 + 3 + 3t^2}{1+t^2} = \frac{4t + 4 + 2t^2}{1+t^2}. \]

### Step 4: Change the Differential \( dx \)
Next, we need to find \( dx \) in terms of \( dt \). The relationship is given by:

\[ dx = \frac{2}{1+t^2} dt. \]

### Step 5: Substitute \( dx \) into the Integral
Now substituting \( dx \) into the integral:

\[ I = \int \frac{2}{1+t^2} \cdot \frac{1+t^2}{4t + 4 + 2t^2} dt = \int \frac{2}{4t + 4 + 2t^2} dt. \]

### Step 6: Factor Out Constants
Factoring out constants in the denominator gives:

\[ I = \int \frac{1}{2t^2 + 4t + 4} dt. \]

### Step 7: Complete the Square
Completing the square for the quadratic in the denominator:

\[ 2t^2 + 4t + 4 = 2(t^2 + 2t + 2) = 2\left((t+1)^2 + 1\right). \]

### Step 8: Substitute Back into the Integral
Thus, we have:

\[ I = \int \frac{1}{2\left((t+1)^2 + 1\right)} dt = \frac{1}{2} \int \frac{1}{(t+1)^2 + 1} dt. \]

### Step 9: Recognize the Standard Integral
This integral is of the form:

\[ \int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C. \]

Here, \( u = t + 1 \) and \( a = 1 \). Therefore,

\[ I = \frac{1}{2} \cdot \frac{1}{1} \tan^{-1}(t + 1) + C = \frac{1}{2} \tan^{-1}(t + 1) + C. \]

### Step 10: Substitute Back for \( t \)
Recall that \( t = \tan\left(\frac{x}{2}\right) \):

\[ I = \frac{1}{2} \tan^{-1}\left(\tan\left(\frac{x}{2}\right) + 1\right) + C. \]

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

\[ \int \frac{dx}{2 \sin x + \cos x + 3} = \frac{1}{2} \tan^{-1}\left(\tan\left(\frac{x}{2}\right) + 1\right) + C. \]