The maximum value of `int _(-pi/2) ^((3pi)/2) sin x. f (x) dx,` subject to the condition `|f (x)| le 5` is `M`, then `M/10` is equal to ______

Answer

To solve the problem, we need to find the maximum value of the integral

\[ M = \int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin x \cdot f(x) \, dx \]

subject to the condition \(|f(x)| \leq 5\).

### Step 1: Understand the bounds of \(f(x)\)
Since \(|f(x)| \leq 5\), we know that:

\[ -5 \leq f(x) \leq 5 \]

This means that the maximum value of \(f(x)\) is 5 and the minimum value is -5.

### Step 2: Analyze the integral
The integral can be split into two parts based on the behavior of \(\sin x\) over the interval \([- \frac{\pi}{2}, \frac{3\pi}{2}]\):

1. From \(-\frac{\pi}{2}\) to \(0\), \(\sin x\) is negative.
2. From \(0\) to \(\frac{\pi}{2}\), \(\sin x\) is positive.
3. From \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\), \(\sin x\) is negative again.

### Step 3: Set \(f(x)\) for maximum contribution
To maximize the integral, we can choose:

- \(f(x) = 5\) in the interval where \(\sin x\) is positive (from \(0\) to \(\frac{\pi}{2}\)).
- \(f(x) = -5\) in the intervals where \(\sin x\) is negative (from \(-\frac{\pi}{2}\) to \(0\) and from \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\)).

### Step 4: Calculate the integral
Now we can express \(M\) as:

\[ M = \int_{-\frac{\pi}{2}}^{0} \sin x \cdot (-5) \, dx + \int_{0}^{\frac{\pi}{2}} \sin x \cdot 5 \, dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin x \cdot (-5) \, dx \]

This simplifies to:

\[ M = -5 \int_{-\frac{\pi}{2}}^{0} \sin x \, dx + 5 \int_{0}^{\frac{\pi}{2}} \sin x \, dx - 5 \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin x \, dx \]

### Step 5: Evaluate each integral
1. **First Integral**:
\[ \int_{-\frac{\pi}{2}}^{0} \sin x \, dx = [-\cos x]_{-\frac{\pi}{2}}^{0} = -\cos(0) + \cos\left(-\frac{\pi}{2}\right) = -1 + 0 = -1 \]
Thus,
\[ -5 \int_{-\frac{\pi}{2}}^{0} \sin x \, dx = -5(-1) = 5 \]

2. **Second Integral**:
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]
Thus,
\[ 5 \int_{0}^{\frac{\pi}{2}} \sin x \, dx = 5(1) = 5 \]

3. **Third Integral**:
\[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin x \, dx = [-\cos x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = -\cos\left(\frac{3\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 0 + 0 = 0 \]
Thus,
\[ -5 \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin x \, dx = -5(0) = 0 \]

### Step 6: Combine results
Now we can combine the results:

\[ M = 5 + 5 + 0 = 10 \]

### Step 7: Find \(M/10\)
Finally, we need to find \(M/10\):

\[ \frac{M}{10} = \frac{10}{10} = 1 \]

Thus, the final answer is:

\[ \boxed{1} \]