`int(1)/(cosx sqrt(cos2x))dx=`

Options

A. `sqrt2[sqrt(tanx)+(1)/(5)tan^(5//2)x]+c`
B. `sqrt2[sqrt(cotx)+(1)/(5)tan^(5//2)x]+c`
C. `sqrt2[sqrt(tanx)-(1)/(5)tan^(5//2)x]+c`
D. `sin^(-1)(tanx)+c`

Answer

To solve the integral \( I = \int \frac{1}{\cos x \sqrt{\cos 2x}} \, dx \), we will follow these steps:

### Step 1: Rewrite \(\cos 2x\)
We know that:
\[ \cos 2x = \cos^2 x - \sin^2 x \]
We can also express \(\cos 2x\) in terms of \(\cos^2 x\):
\[ \cos 2x = 2\cos^2 x - 1 \]
Thus, we can write:
\[ \sqrt{\cos 2x} = \sqrt{2\cos^2 x - 1} \]

### Step 2: Substitute \(\sqrt{\cos 2x}\)
Now, we can substitute this back into the integral:
\[ I = \int \frac{1}{\cos x \sqrt{2\cos^2 x - 1}} \, dx \]

### Step 3: Use the identity for \(\tan x\)
To simplify the integral, we will use the substitution:
\[ \tan x = t \quad \Rightarrow \quad dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \]
Also, we have:
\[ \cos x = \frac{1}{\sqrt{1 + t^2}} \]

### Step 4: Substitute into the integral
Now substituting \(\tan x\) and \(dx\) into the integral:
\[ I = \int \frac{1}{\frac{1}{\sqrt{1 + t^2}} \sqrt{2\left(\frac{1}{1 + t^2}\right)^2 - 1}} \cdot \frac{dt}{1 + t^2} \]
This simplifies to:
\[ I = \int \frac{\sqrt{1 + t^2}}{\sqrt{2 - (1 + t^2)}} \cdot \frac{dt}{1 + t^2} \]

### Step 5: Simplify the expression
The term under the square root simplifies to:
\[ \sqrt{1 + t^2} \cdot \sqrt{1 - \frac{1}{2}(1 + t^2)} = \sqrt{1 + t^2} \cdot \sqrt{\frac{1}{2}(1 - t^2)} \]
Thus, we can write:
\[ I = \int \frac{dt}{\sqrt{1 - t^2}} \]

### Step 6: Recognize the standard integral
The integral \(\int \frac{dt}{\sqrt{1 - t^2}}\) is a standard integral that evaluates to:
\[ \sin^{-1}(t) + C \]

### Step 7: Substitute back for \(t\)
Since \(t = \tan x\), we have:
\[ I = \sin^{-1}(\tan x) + C \]

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

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