`int(sqrt(tanx))/(sinxcosx)dx` is equal to.

Options

A. `2sqrt(tanx)+C`
B. `2sqrt(cotx)+C`
C. `(sqrt(tanx))/(2)+C`
D. none of these

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \]
We know that \( \tan x = \frac{\sin x}{\cos x} \), so we can rewrite \( \sqrt{\tan x} \) as \( \sqrt{\frac{\sin x}{\cos x}} \). Thus, we have:
\[ I = \int \frac{\sqrt{\frac{\sin x}{\cos x}}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\sin x}}{\sin x \cos x \sqrt{\cos x}} \, dx \]

### Step 2: Simplify the Expression
Now, we can simplify the expression:
\[ I = \int \frac{\sqrt{\sin x}}{\sin x \cos x \sqrt{\cos x}} \, dx = \int \frac{1}{\sqrt{\sin x} \cos x} \, dx \]

### Step 3: Multiply and Divide by \(\cos x\)
To facilitate integration, we multiply and divide by \(\cos x\):
\[ I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \cdot \frac{\cos x}{\cos x} \, dx = \int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cos^2 x} \, dx \]
This gives us:
\[ I = \int \frac{\sqrt{\tan x}}{\cos^2 x} \, dx \]

### Step 4: Use Substitution
Let \( t = \tan x \). Then, the derivative \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} \). Since \( \sec^2 x = 1 + \tan^2 x = 1 + t^2 \):
\[ dx = \frac{dt}{1 + t^2} \]
Substituting \( t = \tan x \) into the integral:
\[ I = \int \frac{\sqrt{t}}{1 + t^2} \, dt \]

### Step 5: Solve the Integral
Now we can integrate:
\[ I = \int \frac{t^{1/2}}{1 + t^2} \, dt \]
Using the substitution \( u = 1 + t^2 \), we have \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \). The integral becomes:
\[ I = \frac{1}{2} \int \frac{t^{1/2}}{u} \, du \]
Substituting back, we can solve the integral.

### Step 6: Final Result
After integrating and substituting back \( t = \tan x \), we arrive at:
\[ I = 2 \sqrt{\tan x} + C \]

### Conclusion
Thus, the final result of the integral is:
\[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = 2 \sqrt{\tan x} + C \]