Compute the integrals :
`int_(pi//4)^(pi//3) (x dx)/( sin^(2) x)`

Answer

To compute the integral

\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x \, dx}{\sin^2 x}, \]

we will use integration by parts.

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = x \) (which means \( du = dx \))
- \( dv = \frac{dx}{\sin^2 x} \) (which means we need to find \( v \))

### Step 2: Find \( v \)

To find \( v \), we need to integrate \( dv \):

\[ v = \int \frac{dx}{\sin^2 x} = -\cot x. \]

### Step 3: Apply integration by parts

Using the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du, \]

we have:

\[ I = \left[-x \cot x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} - \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (-\cot x) \, dx. \]

### Step 4: Evaluate the boundary term

Now we evaluate the boundary term:

\[ \left[-x \cot x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -\left(\frac{\pi}{3} \cot\left(\frac{\pi}{3}\right) - \frac{\pi}{4} \cot\left(\frac{\pi}{4}\right)\right). \]

We know that:

\[ \cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}, \quad \cot\left(\frac{\pi}{4}\right) = 1. \]

Thus, substituting these values:

\[ = -\left(\frac{\pi}{3} \cdot \frac{1}{\sqrt{3}} - \frac{\pi}{4} \cdot 1\right) = -\left(\frac{\pi}{3\sqrt{3}} - \frac{\pi}{4}\right). \]

### Step 5: Simplify the expression

Now, we need to simplify:

\[ -\left(\frac{\pi}{3\sqrt{3}} - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{\pi}{3\sqrt{3}}. \]

### Step 6: Evaluate the remaining integral

Now we need to evaluate the integral:

\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x \, dx. \]

The integral of \( \cot x \) is \( \ln(\sin x) \):

\[ \int \cot x \, dx = \ln(\sin x). \]

Thus,

\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x \, dx = \left[\ln(\sin x)\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \ln(\sin(\frac{\pi}{3})) - \ln(\sin(\frac{\pi}{4})). \]

We know:

\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}. \]

Thus,

\[ = \ln\left(\frac{\sqrt{3}}{2}\right) - \ln\left(\frac{1}{\sqrt{2}}\right) = \ln\left(\frac{\sqrt{3}}{2} \cdot \sqrt{2}\right) = \ln\left(\frac{\sqrt{6}}{2}\right). \]

### Step 7: Combine results

Combining everything, we have:

\[ I = \left(\frac{\pi}{4} - \frac{\pi}{3\sqrt{3}}\right) + \ln\left(\frac{\sqrt{6}}{2}\right). \]

### Final Answer

Thus, the value of the integral is:

\[ I = \frac{\pi}{4} - \frac{\pi}{3\sqrt{3}} + \ln\left(\frac{\sqrt{6}}{2}\right). \]