`f(x)=int_0^x(e^t-1)(t-1)(sint-cost)sin t dt ,AAx in (-pi/2, 2pi),` then `f(x)` is decreasing in :

Options

A. `(-pi/2,0) cup (pi/4,1) cup (pi,pi/4)`
B. `(-pi/2,pi/4) cup (1, pi) cup ((5pi)/(4), 2pi)`
C. `(pi/4,1) cup (pi,(5pi)/(4))`
D. `(0,pi/4) cup (1, pi) cup ((5pi)/(4), 2pi)`

Answer

To determine where the function \( f(x) \) is decreasing, we need to analyze the derivative \( f'(x) \). The function is given by:

\[ f(x) = \int_0^x (e^t - 1)(t - 1)(\sin t - \cos t) \sin t \, dt \]

### Step 1: Differentiate \( f(x) \)

Using the Fundamental Theorem of Calculus, we differentiate \( f(x) \):

\[ f'(x) = (e^x - 1)(x - 1)(\sin x - \cos x) \sin x \]

### Step 2: Analyze the derivative \( f'(x) \)

To find where \( f(x) \) is decreasing, we need to determine where \( f'(x) < 0 \). This involves analyzing the sign of each factor in \( f'(x) \):

1. **Factor 1: \( e^x - 1 \)**
- \( e^x - 1 = 0 \) when \( x = 0 \)
- \( e^x - 1 > 0 \) for \( x > 0 \)
- \( e^x - 1 < 0 \) for \( x < 0 \)

2. **Factor 2: \( x - 1 \)**
- \( x - 1 = 0 \) when \( x = 1 \)
- \( x - 1 > 0 \) for \( x > 1 \)
- \( x - 1 < 0 \) for \( x < 1 \)

3. **Factor 3: \( \sin x - \cos x \)**
- \( \sin x - \cos x = 0 \) when \( \tan x = 1 \)
- This occurs at \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \) within the interval \( (-\frac{\pi}{2}, 2\pi) \).

4. **Factor 4: \( \sin x \)**
- \( \sin x = 0 \) at \( x = 0, \pi, 2\pi \)
- \( \sin x > 0 \) in the intervals \( (0, \pi) \) and \( (2\pi, 2\pi) \)
- \( \sin x < 0 \) in the intervals \( (-\frac{\pi}{2}, 0) \) and \( (\pi, 2\pi) \)

### Step 3: Determine intervals for \( f'(x) < 0 \)

Now we will analyze the sign of \( f'(x) \) in the intervals defined by the critical points \( x = 0, 1, \frac{\pi}{4}, \pi, \frac{5\pi}{4}, 2\pi \):

1. **Interval \( (-\frac{\pi}{2}, 0) \)**
- \( e^x - 1 < 0 \)
- \( x - 1 < 0 \)
- \( \sin x - \cos x < 0 \)
- \( \sin x < 0 \)
- Thus, \( f'(x) > 0 \)

2. **Interval \( (0, \frac{\pi}{4}) \)**
- \( e^x - 1 > 0 \)
- \( x - 1 < 0 \)
- \( \sin x - \cos x < 0 \)
- \( \sin x > 0 \)
- Thus, \( f'(x) < 0 \)

3. **Interval \( (\frac{\pi}{4}, 1) \)**
- \( e^x - 1 > 0 \)
- \( x - 1 < 0 \)
- \( \sin x - \cos x > 0 \)
- \( \sin x > 0 \)
- Thus, \( f'(x) < 0 \)

4. **Interval \( (1, \pi) \)**
- \( e^x - 1 > 0 \)
- \( x - 1 > 0 \)
- \( \sin x - \cos x < 0 \)
- \( \sin x > 0 \)
- Thus, \( f'(x) < 0 \)

5. **Interval \( (\pi, \frac{5\pi}{4}) \)**
- \( e^x - 1 > 0 \)
- \( x - 1 > 0 \)
- \( \sin x - \cos x < 0 \)
- \( \sin x < 0 \)
- Thus, \( f'(x) < 0 \)

6. **Interval \( (\frac{5\pi}{4}, 2\pi) \)**
- \( e^x - 1 > 0 \)
- \( x - 1 > 0 \)
- \( \sin x - \cos x < 0 \)
- \( \sin x < 0 \)
- Thus, \( f'(x) < 0 \)

### Conclusion

The function \( f(x) \) is decreasing in the intervals:

- \( (0, \frac{\pi}{4}) \)
- \( (\frac{\pi}{4}, 1) \)
- \( (1, \pi) \)
- \( (\pi, \frac{5\pi}{4}) \)
- \( (\frac{5\pi}{4}, 2\pi) \)