`intsqrt(1+cos e cx)dx e q u a l s` `2sin^(-1)sqrt(sinx)+c` (b) `sqrt(2)cos^(-1)sqrt(cosx)+c` `c-2sin^(-1)(1-2sinx)` `cos^(-1)(1-2sinx)+c`

Options

A. `2sin^(-1)sqrt(sinx)+c`
B. `sqrt(2)cos^(-1)sqrt(cosx)+c`
C. `c-2sin^(-1)(1-2sinx)`
D. `cos^(-1)(1-2sinx)+c`

Explanation

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \sqrt{1 + \cos x} \, dx \]
Using the identity \(1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)\), we can rewrite the integral:
\[ \int \sqrt{2 \cos^2\left(\frac{x}{2}\right)} \, dx = \int \sqrt{2} \cos\left(\frac{x}{2}\right) \, dx \]

### Step 2: Factor Out the Constant
We can factor out the constant \(\sqrt{2}\):
\[ \sqrt{2} \int \cos\left(\frac{x}{2}\right) \, dx \]

### Step 3: Perform the Integration
To integrate \(\cos\left(\frac{x}{2}\right)\), we use the substitution \(u = \frac{x}{2}\), which gives \(du = \frac{1}{2} dx\) or \(dx = 2 du\):
\[ \sqrt{2} \int \cos(u) \cdot 2 \, du = 2\sqrt{2} \int \cos(u) \, du \]
The integral of \(\cos(u)\) is \(\sin(u)\):
\[ 2\sqrt{2} \sin(u) + C \]

### Step 4: Substitute Back
Now we substitute back \(u = \frac{x}{2}\):
\[ 2\sqrt{2} \sin\left(\frac{x}{2}\right) + C \]

### Step 5: Final Simplification
We can express \(\sin\left(\frac{x}{2}\right)\) in terms of \(\sqrt{\sin x}\) using the double angle identity:
\[ \sin\left(\frac{x}{2}\right) = \sqrt{\sin x} \]
Thus, our final answer becomes:
\[ 2 \sin^{-1}(\sqrt{\sin x}) + C \]

### Conclusion
The final result of the integral is:
\[ \int \sqrt{1 + \cos x} \, dx = 2 \sin^{-1}(\sqrt{\sin x}) + C \]