`int(sqrt((cosx)/(x))-sqrt((x)/(cosx))sinx)dx` equals

Options

A. `-sqrt(xcosx)+C`
B. `sqrt(xsinx)+C`
C. `2sqrt(xcos x)+C`
D. `C-2sqrt(xcosx)`

Explanation

To solve the integral

\[ \int \left( \sqrt{\frac{\cos x}{x}} - \sqrt{\frac{x}{\cos x}} \sin x \right) dx, \]

we will follow a systematic approach.

### Step 1: Combine the terms under a common denominator

We start by rewriting the integrand with a common denominator:

\[ \sqrt{\frac{\cos x}{x}} - \sqrt{\frac{x}{\cos x}} \sin x = \frac{\sqrt{\cos x} \cdot \sqrt{\cos x} - \sqrt{x} \cdot \sin x \cdot \sqrt{x}}{\sqrt{x \cos x}} = \frac{\cos x - x \sin x}{\sqrt{x \cos x}}. \]

### Step 2: Rewrite the integral

Now we can rewrite the integral as:

\[ \int \frac{\cos x - x \sin x}{\sqrt{x \cos x}} \, dx. \]

### Step 3: Substitution

Next, we will use the substitution \( t = x \cos x \). Then, we need to find \( dt \):

Using the product rule, we differentiate:

\[ dt = \cos x \, dx - x \sin x \, dx = (\cos x - x \sin x) \, dx. \]

Thus, we have:

\[ \cos x - x \sin x \, dx = dt. \]

### Step 4: Substitute in the integral

Now, substituting back into the integral, we have:

\[ \int \frac{dt}{\sqrt{t}}. \]

### Step 5: Integrate

The integral of \( \frac{1}{\sqrt{t}} \) is:

\[ \int t^{-1/2} \, dt = 2t^{1/2} + C. \]

### Step 6: Substitute back for \( t \)

Now we substitute back \( t = x \cos x \):

\[ 2\sqrt{x \cos x} + C. \]

### Final Answer

Thus, the value of the integral is:

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

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