`int_(pi//3)^(pi//2) (sqrt(1+cosx))/((1-cosx)^(5//2))\ dx`

Explanation

To solve the integral

\[ I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{\frac{5}{2}}} \, dx, \]

we will use trigonometric identities and substitution.

### Step 1: Rewrite the integrand using trigonometric identities

We know that

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

Thus, we can express \(1 + \cos x\) as:

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

Now substituting these into the integral, we have:

\[ I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{2 \cos^2\left(\frac{x}{2}\right)}}{(2 \sin^2\left(\frac{x}{2}\right))^{\frac{5}{2}}} \, dx. \]

### Step 2: Simplify the integrand

The integral simplifies to:

\[ I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{2} \cos\left(\frac{x}{2}\right)}{(2^{\frac{5}{2}} \sin^5\left(\frac{x}{2}\right))} \, dx = \frac{\sqrt{2}}{4\sqrt{2}} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} \, dx. \]

This simplifies to:

\[ I = \frac{1}{4} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} \, dx. \]

### Step 3: Substitution

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

\[ \frac{dx}{dt} = \frac{2 \cos\left(\frac{x}{2}\right)}{1} \implies dx = \frac{2 \cos\left(\frac{x}{2}\right)}{dt}. \]

Now, we need to change the limits of integration:

- When \(x = \frac{\pi}{3}\), \(t = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\).
- When \(x = \frac{\pi}{2}\), \(t = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\).

Now, substituting into the integral, we have:

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

### Step 4: Integrate

The integral of \(t^{-5}\) is:

\[ \int t^{-5} \, dt = \frac{t^{-4}}{-4} = -\frac{1}{4t^4}. \]

Now, applying the limits:

\[ I = \frac{1}{2} \left[-\frac{1}{4t^4}\right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} = \frac{1}{2} \left(-\frac{1}{4\left(\frac{1}{\sqrt{2}}\right)^4} + \frac{1}{4\left(\frac{1}{2}\right)^4}\right). \]

Calculating the values:

\[ = \frac{1}{2} \left(-\frac{1}{4 \cdot \frac{1}{4}} + \frac{1}{4 \cdot \frac{1}{16}}\right) = \frac{1}{2} \left(-1 + 4\right) = \frac{1}{2} \cdot 3 = \frac{3}{2}. \]

### Final Answer

Thus, the value of the integral is:

\[ I = \frac{3}{2}. \]