`int_0^pi sqrt(1+cosx)/(1-cosx)^(5//2) dx`

Answer

To solve the integral

\[ I = \int_0^{\pi} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} \, dx, \]

we can follow these steps:

### Step 1: Simplify the integrand

We know that:

\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \quad \text{and} \quad 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right). \]

Substituting these into the integral gives:

\[ I = \int_0^{\pi} \frac{\sqrt{2 \cos^2\left(\frac{x}{2}\right)}}{(2 \sin^2\left(\frac{x}{2}\right))^{5/2}} \, dx. \]

### Step 2: Rewrite the integral

This simplifies to:

\[ I = \int_0^{\pi} \frac{\sqrt{2} \cos\left(\frac{x}{2}\right)}{(2^{5/2} \sin^5\left(\frac{x}{2}\right))} \, dx. \]

We can factor out constants:

\[ I = \frac{\sqrt{2}}{4\sqrt{2}} \int_0^{\pi} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} \, dx = \frac{1}{4} \int_0^{\pi} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} \, dx. \]

### Step 3: Change of variables

Let \( t = \sin\left(\frac{x}{2}\right) \). Then, we have:

\[ dx = \frac{2}{\sqrt{1 - t^2}} \, dt. \]

Also, when \( x = 0 \), \( t = 0 \) and when \( x = \pi \), \( t = 1 \). Thus, the integral becomes:

\[ I = \frac{1}{4} \int_0^1 \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} \cdot \frac{2}{\sqrt{1 - t^2}} \, dt. \]

### Step 4: Express \(\cos\left(\frac{x}{2}\right)\) in terms of \(t\)

Using the identity \( \cos\left(\frac{x}{2}\right) = \sqrt{1 - \sin^2\left(\frac{x}{2}\right)} = \sqrt{1 - t^2} \):

\[ I = \frac{1}{4} \int_0^1 \frac{\sqrt{1 - t^2}}{t^5} \cdot \frac{2}{\sqrt{1 - t^2}} \, dt = \frac{1}{2} \int_0^1 t^{-5} \, dt. \]

### Step 5: Evaluate the integral

Now we can evaluate:

\[ I = \frac{1}{2} \int_0^1 t^{-5} \, dt = \frac{1}{2} \left[ \frac{t^{-4}}{-4} \right]_0^1. \]

This gives:

\[ = \frac{1}{2} \left( -\frac{1}{4} (1 - 0) \right) = -\frac{1}{8}. \]

### Final Answer

Thus, the value of the integral is:

\[ I = -\frac{1}{8}. \]