Evaluate `int(sqrt(tanx)+sqrt(cotx))dx`

Explanation

To evaluate the integral \( \int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting \( \tan x \) and \( \cot x \) in terms of sine and cosine:
\[ \sqrt{\tan x} = \sqrt{\frac{\sin x}{\cos x}} \quad \text{and} \quad \sqrt{\cot x} = \sqrt{\frac{\cos x}{\sin x}} \]
Thus, we can express the integral as:
\[ \int \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) \, dx \]

### Step 2: Combine the terms
Next, we can combine the two square roots over a common denominator:
\[ \sqrt{\tan x} + \sqrt{\cot x} = \frac{\sqrt{\sin x} \cdot \sqrt{\sin x} + \sqrt{\cos x} \cdot \sqrt{\cos x}}{\sqrt{\sin x \cos x}} = \frac{\sqrt{\sin^2 x + \cos^2 x}}{\sqrt{\sin x \cos x}} \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sqrt{\tan x} + \sqrt{\cot x} = \frac{1}{\sqrt{\sin x \cos x}} \]
Thus, the integral becomes:
\[ \int \frac{1}{\sqrt{\sin x \cos x}} \, dx \]

### Step 3: Simplify using a trigonometric identity
We can use the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \):
\[ \sqrt{\sin x \cos x} = \sqrt{\frac{1}{2} \sin(2x)} = \frac{1}{\sqrt{2}} \sqrt{\sin(2x)} \]
So, we can rewrite the integral as:
\[ \int \frac{1}{\frac{1}{\sqrt{2}} \sqrt{\sin(2x)}} \, dx = \sqrt{2} \int \frac{1}{\sqrt{\sin(2x)}} \, dx \]

### Step 4: Use substitution
Let \( t = \sin(2x) \), then \( dt = 2\cos(2x) \, dx \) or \( dx = \frac{dt}{2\cos(2x)} \). We also need to express \( \cos(2x) \) in terms of \( t \):
\[ \cos(2x) = \sqrt{1 - t^2} \]
Substituting these into the integral gives:
\[ \sqrt{2} \int \frac{1}{\sqrt{t}} \cdot \frac{dt}{2\sqrt{1 - t^2}} = \frac{\sqrt{2}}{2} \int \frac{1}{\sqrt{t(1 - t^2)}} \, dt \]

### Step 5: Solve the integral
This integral can be solved using standard integral tables or known results:
\[ \int \frac{1}{\sqrt{t(1 - t^2)}} \, dt = \sin^{-1}(\sqrt{t}) + C \]
Thus, we have:
\[ \frac{\sqrt{2}}{2} \left( \sin^{-1}(\sqrt{\sin(2x)}) + C \right) \]

### Final Answer
Therefore, the evaluated integral is:
\[ \frac{\sqrt{2}}{2} \sin^{-1}(\sqrt{\sin(2x)}) + C \]