Evaluate :
`int sqrt(tan x) (1+ tan^(2)x)dx`

Answer

To evaluate the integral

\[ \int \sqrt{\tan x} (1 + \tan^2 x) \, dx, \]

we can follow these steps:

### Step 1: Simplify the integrand
We know that

\[ 1 + \tan^2 x = \sec^2 x. \]

Thus, we can rewrite the integral as:

\[ \int \sqrt{\tan x} \sec^2 x \, dx. \]

### Step 2: Substitute \( t = \tan x \)
Let \( t = \tan x \). Then, the derivative of \( \tan x \) is:

\[ \frac{dt}{dx} = \sec^2 x \implies dx = \frac{dt}{\sec^2 x}. \]

### Step 3: Substitute in the integral
Substituting \( t \) into the integral gives:

\[ \int \sqrt{t} \sec^2 x \cdot \frac{dt}{\sec^2 x} = \int \sqrt{t} \, dt. \]

### Step 4: Evaluate the integral
Now we can evaluate the integral:

\[ \int \sqrt{t} \, dt = \int t^{1/2} \, dt. \]

Using the power rule for integration:

\[ \int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C, \]

we have:

\[ \int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} + C = \frac{2}{3} t^{3/2} + C. \]

### Step 5: Substitute back for \( t \)
Recalling that \( t = \tan x \), we substitute back:

\[ \frac{2}{3} \tan^{3/2} x + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ \int \sqrt{\tan x} (1 + \tan^2 x) \, dx = \frac{2}{3} \tan^{3/2} x + C. \]

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