`int(cosx)/(sqrt(cos2x))dx=`

Options

A. `(-1)/(sqrt2)log[sqrt2sinx+sqrt(cos2x)]+c`
B. `(-1)/(sqrt2)sin^(-1)(sqrt2 cos x)+c`
C. `(1)/(sqrt2)sin^(-1)(sqrt2 sinx)+c`
D. `(1)/(sqrt2)sin^(-1)(sqrt2sinx)+c`

Answer

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

### Step 1: Rewrite the Integral
We know that \( \cos 2x = 1 - 2\sin^2 x \). Therefore, we can rewrite the integral as:
\[ \int \frac{\cos x}{\sqrt{1 - 2\sin^2 x}} \, dx \]

### Step 2: Use Substitution
Let \( t = \sin x \). Then, we have:
\[ dt = \cos x \, dx \quad \text{or} \quad dx = \frac{dt}{\cos x} \]
Substituting this into the integral gives:
\[ \int \frac{\cos x}{\sqrt{1 - 2t^2}} \cdot \frac{dt}{\cos x} = \int \frac{1}{\sqrt{1 - 2t^2}} \, dt \]

### Step 3: Factor Out Constants
We can factor out constants from the square root:
\[ \int \frac{1}{\sqrt{1 - 2t^2}} \, dt = \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\frac{1}{2} - t^2}} \, dt \]

### Step 4: Recognize the Integral Form
The integral \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx \) has a known result:
\[ \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + C \]
In our case, \( a^2 = \frac{1}{2} \) so \( a = \frac{1}{\sqrt{2}} \).

### Step 5: Apply the Formula
Using the formula, we have:
\[ \int \frac{1}{\sqrt{\frac{1}{2} - t^2}} \, dt = \sin^{-1} \left( \frac{t}{\frac{1}{\sqrt{2}}} \right) + C = \sin^{-1}(\sqrt{2} t) + C \]

### Step 6: Substitute Back
Now substitute back \( t = \sin x \):
\[ \sin^{-1}(\sqrt{2} \sin x) + C \]

### Step 7: Multiply by the Constant
Don’t forget the factor we took out earlier:
\[ \frac{1}{\sqrt{2}} \left( \sin^{-1}(\sqrt{2} \sin x) + C \right) \]
Thus, the final result is:
\[ \frac{1}{\sqrt{2}} \sin^{-1}(\sqrt{2} \sin x) + C \]

### Final Answer
The final answer to the integral is:
\[ \frac{1}{\sqrt{2}} \sin^{-1}(\sqrt{2} \sin x) + C \]