`int(sinx+cosx)/(sqrt(1+sin2x))dx`

Options

A. `x^2/5 + C`
B. `x + C`
C. `2x + C`
D. `x ^2+ C`

Explanation

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

### Step 1: Simplify the Denominator
We know that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite the denominator:

\[ 1 + \sin 2x = 1 + 2 \sin x \cos x \]

### Step 2: Use Trigonometric Identity
We can also use the identity \( \sin^2 x + \cos^2 x = 1 \). Thus, we can rewrite the expression under the square root:

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

### Step 3: Substitute the Denominator
Now, substituting this back into the integral gives us:

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

Since \( \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x| \), we can simplify the integral:

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

### Step 4: Determine the Sign
The expression \( \sin x + \cos x \) can be positive or negative depending on the value of \( x \). Therefore, we consider two cases:

1. **Case 1**: When \( \sin x + \cos x \geq 0 \)
\[ \int 1 \, dx = x + C \]

2. **Case 2**: When \( \sin x + \cos x < 0 \)
\[ \int -1 \, dx = -x + C \]

### Step 5: Combine Results
Thus, the integral can be expressed as:

\[ \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx = \begin{cases} x + C & \text{if } \sin x + \cos x \geq 0 \\ -x + C & \text{if } \sin x + \cos x < 0 \end{cases} \]