Evaluate `int(sqrt(tanx))/(sinx cosx)dx`

Explanation

To evaluate the integral \( \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \), we can 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 \( \tan x = \frac{\sin x}{\cos x} \). Thus, we can rewrite \( \sqrt{\tan x} \) as:
\[ \sqrt{\tan x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \]
Now, substituting this into the integral gives:
\[ I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\sin x}}{\sin x \cos x \sqrt{\cos x}} \, dx = \int \frac{1}{\sin x \sqrt{\cos x}} \, dx \]

### Step 2: Simplify the Integral
Next, we can simplify the integral further. We can divide both the numerator and denominator by \( \cos^2 x \):
\[ I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\tan x}}{\tan x \sin^2 x} \, dx \]
This simplifies to:
\[ I = \int \frac{1}{\tan x} \cdot \sin^2 x \, dx \]

### Step 3: Substitution
Now, we perform the substitution \( t = \tan x \). Then, we know:
\[ \frac{dt}{dx} = \sec^2 x \quad \text{or} \quad dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \]
Also, we have:
\[ \sin^2 x = \frac{\tan^2 x}{1 + \tan^2 x} = \frac{t^2}{1 + t^2} \]
Thus, the integral becomes:
\[ I = \int \frac{1}{t} \cdot \frac{t^2}{1 + t^2} \cdot \frac{dt}{1 + t^2} = \int \frac{t}{(1 + t^2)^2} \, dt \]

### Step 4: Solve the Integral
Now, we can integrate:
\[ I = \int \frac{t}{(1 + t^2)^2} \, dt \]
Using the substitution \( u = 1 + t^2 \), we have \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \). The integral becomes:
\[ I = \frac{1}{2} \int \frac{1}{u^2} \, du = -\frac{1}{2u} + C = -\frac{1}{2(1 + t^2)} + C \]

### Step 5: Back Substitute
Now, substituting back \( t = \tan x \):
\[ I = -\frac{1}{2(1 + \tan^2 x)} + C \]
Using the identity \( 1 + \tan^2 x = \sec^2 x \):
\[ I = -\frac{1}{2 \sec^2 x} + C = -\frac{1}{2} \cos^2 x + C \]

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