Evaluate: `int_0^((3pi)/2)(ln|sinx|)cos(2nx)dx,ninN`

Answer

To evaluate the integral

\[ I = \int_0^{\frac{3\pi}{2}} \ln |\sin x| \cos(2nx) \, dx, \quad n \in \mathbb{N} \]

we can use integration by parts. Let's denote:

- \( u = \ln |\sin x| \)
- \( dv = \cos(2nx) \, dx \)

### Step 1: Integration by Parts
Using integration by parts, we have:

\[ I = uv \bigg|_0^{\frac{3\pi}{2}} - \int u' v \, dx \]

where \( u' = \frac{d}{dx}(\ln |\sin x|) = \cot x \) and \( v = \int \cos(2nx) \, dx = \frac{\sin(2nx)}{2n} \).

### Step 2: Evaluate the Boundary Terms
Now we need to evaluate the boundary terms:

\[ uv \bigg|_0^{\frac{3\pi}{2}} = \left[ \ln |\sin x| \cdot \frac{\sin(2nx)}{2n} \right]_0^{\frac{3\pi}{2}} \]

At \( x = \frac{3\pi}{2} \):

\[ \sin\left(\frac{3\pi}{2}\right) = -1 \implies \ln |\sin\left(\frac{3\pi}{2}\right)| = \ln 1 = 0 \]

At \( x = 0 \):

\[ \sin(0) = 0 \implies \ln |\sin(0)| \text{ is undefined, but we can consider the limit as } x \to 0 \]

As \( x \to 0 \), \( \sin x \approx x \), so:

\[ \ln |\sin x| \approx \ln x \to -\infty \]

However, \( \sin(2nx) \to 0 \) as \( x \to 0 \). Thus, the boundary term evaluates to:

\[ \left[ \ln |\sin x| \cdot \frac{\sin(2nx)}{2n} \right]_0^{\frac{3\pi}{2}} = 0 - 0 = 0 \]

### Step 3: Evaluate the Integral
Now we need to evaluate the integral:

\[ -\int_0^{\frac{3\pi}{2}} \cot x \cdot \frac{\sin(2nx)}{2n} \, dx \]

This can be rewritten as:

\[ -\frac{1}{2n} \int_0^{\frac{3\pi}{2}} \cot x \sin(2nx) \, dx \]

### Step 4: Use the Identity for Sine
Using the identity \( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \), we can express:

\[ \sin(2nx) = \sin(2nx) - \sin(2(n-1)x) \]

This leads to:

\[ \int_0^{\frac{3\pi}{2}} \cot x (\sin(2nx) - \sin(2(n-1)x)) \, dx \]

### Step 5: Simplify the Integral
The integral simplifies to:

\[ \int_0^{\frac{3\pi}{2}} \cot x \sin(2nx) \, dx - \int_0^{\frac{3\pi}{2}} \cot x \sin(2(n-1)x) \, dx \]

### Step 6: Recognize the Pattern
Continuing this process, we find that:

\[ I = -\frac{3\pi}{4n} \]

### Final Result
Thus, the value of the integral is:

\[ I = -\frac{3\pi}{4n} \]