Evaluate: `int_0^(pi//4) secx. sqrt((1-sinx)/(1+sinx)) dx`

Answer

To evaluate the integral \( I = \int_0^{\frac{\pi}{4}} \sec x \sqrt{\frac{1 - \sin x}{1 + \sin x}} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^{\frac{\pi}{4}} \sec x \sqrt{\frac{1 - \sin x}{1 + \sin x}} \, dx \]

### Step 2: Simplify the Square Root
We can simplify the square root:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} = \frac{\sqrt{1 - \sin x}}{\sqrt{1 + \sin x}} \]
Thus, we rewrite the integral as:
\[ I = \int_0^{\frac{\pi}{4}} \sec x \cdot \frac{\sqrt{1 - \sin x}}{\sqrt{1 + \sin x}} \, dx \]

### Step 3: Multiply and Divide by \(1 - \sin x\)
To further simplify, we multiply and divide by \(1 - \sin x\):
\[ I = \int_0^{\frac{\pi}{4}} \sec x \cdot \frac{(1 - \sin x)}{\sqrt{(1 - \sin x)(1 + \sin x)}} \, dx \]
This gives us:
\[ I = \int_0^{\frac{\pi}{4}} \sec x \cdot \frac{(1 - \sin x)}{\sqrt{1 - \sin^2 x}} \, dx \]
Since \(1 - \sin^2 x = \cos^2 x\), we have:
\[ \sqrt{1 - \sin^2 x} = \cos x \]
So the integral becomes:
\[ I = \int_0^{\frac{\pi}{4}} \sec x \cdot \frac{(1 - \sin x)}{\cos x} \, dx \]

### Step 4: Simplify the Expression
This can be simplified to:
\[ I = \int_0^{\frac{\pi}{4}} \sec^2 x (1 - \sin x) \, dx \]
This can be split into two integrals:
\[ I = \int_0^{\frac{\pi}{4}} \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \sec^2 x \sin x \, dx \]

### Step 5: Evaluate the First Integral
The first integral is:
\[ \int \sec^2 x \, dx = \tan x \]
Evaluating from \(0\) to \(\frac{\pi}{4}\):
\[ \left[ \tan x \right]_0^{\frac{\pi}{4}} = \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1 \]

### Step 6: Evaluate the Second Integral
For the second integral, we use integration by parts:
Let \(u = \sin x\) and \(dv = \sec^2 x \, dx\). Then \(du = \cos x \, dx\) and \(v = \tan x\).
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int \sin x \sec^2 x \, dx = \sin x \tan x \bigg|_0^{\frac{\pi}{4}} - \int \tan x \cos x \, dx \]
Evaluating \( \sin x \tan x \) from \(0\) to \(\frac{\pi}{4}\):
\[ \sin\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) - \sin(0) \tan(0) = \frac{\sqrt{2}}{2} \cdot 1 - 0 = \frac{\sqrt{2}}{2} \]
The second integral simplifies to:
\[ \int \tan x \cos x \, dx = \int \sin x \, dx = -\cos x \bigg|_0^{\frac{\pi}{4}} = -\left(-\frac{\sqrt{2}}{2} + 1\right) = 1 - \frac{\sqrt{2}}{2} \]

### Step 7: Combine the Results
Putting it all together:
\[ I = 1 - \left( \frac{\sqrt{2}}{2} - \left(1 - \frac{\sqrt{2}}{2}\right) \right) \]
Thus:
\[ I = 1 - \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} = 2 - \sqrt{2} \]

### Final Result
The value of the integral is:
\[ \boxed{2 - \sqrt{2}} \]