`int(sqrt(tanx))/(sin2x)dx=`

Options

A. `2sqrt(sin2x)+c`
B. `sqrt(tanx)+c`
C. `sqrt(sinx+cosx)+c`
D. `sqrt(cotx)+c`

Answer

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

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

### Step 2: Express \( \tan x \) in terms of \( \sin x \) and \( \cos x \)
Recall that \( \tan x = \frac{\sin x}{\cos x} \). Thus, we can express \( \sqrt{\tan x} \) as:
\[ \sqrt{\tan x} = \sqrt{\frac{\sin x}{\cos x}} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \]

### Step 3: Substitute into the integral
Substituting this into our integral gives:
\[ \int \frac{\frac{\sqrt{\sin x}}{\sqrt{\cos x}}}{2 \sin x \cos x} \, dx = \int \frac{\sqrt{\sin x}}{2 \sin x \cos x \sqrt{\cos x}} \, dx \]
This simplifies to:
\[ \int \frac{1}{2 \sqrt{\cos x} \sin x} \, dx \]

### Step 4: Use substitution
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} \). Also, we know that \( \sin x = \frac{t}{\sqrt{1+t^2}} \) and \( \cos x = \frac{1}{\sqrt{1+t^2}} \).

### Step 5: Substitute \( \sin x \) and \( \cos x \)
Substituting these into the integral:
\[ \int \frac{1}{2 \sqrt{\frac{1}{1+t^2}} \cdot \frac{t}{\sqrt{1+t^2}}} \cdot \frac{dt}{\sec^2 x} \]
This becomes:
\[ \int \frac{\sqrt{1+t^2}}{2t} \cdot \frac{dt}{1+t^2} = \int \frac{1}{2t} \, dt \]

### Step 6: Integrate
Now, we can integrate:
\[ \int \frac{1}{2t} \, dt = \frac{1}{2} \ln |t| + C \]

### Step 7: Substitute back for \( t \)
Recall that \( t = \tan x \):
\[ \frac{1}{2} \ln |\tan x| + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \frac{\sqrt{\tan x}}{\sin 2x} \, dx = \frac{1}{2} \ln |\tan x| + C \]