`int(sin((pi)/(4)-x)dx)/(2+sin2x)=A tan^(-1)(f(x))+B`, where A, B are contants. Then the range of Af(x) is

Options

A. `[-1,1]`
B. `[-sqrt2,sqrt2]`
C. `[0,1]`
D. `[-1,0]`

Explanation

To solve the integral \(\int \frac{\sin\left(\frac{\pi}{4} - x\right)dx}{2 + \sin(2x)}\), we will follow these steps:

### Step 1: Simplify the integrand
Using the sine subtraction formula, we can rewrite \(\sin\left(\frac{\pi}{4} - x\right)\):
\[ \sin\left(\frac{\pi}{4} - x\right) = \sin\left(\frac{\pi}{4}\right)\cos(x) - \cos\left(\frac{\pi}{4}\right)\sin(x) = \frac{1}{\sqrt{2}}\cos(x) - \frac{1}{\sqrt{2}}\sin(x) \]
Thus, the integral becomes:
\[ \int \frac{\frac{1}{\sqrt{2}}(\cos(x) - \sin(x))dx}{2 + \sin(2x)} \]

### Step 2: Rewrite \(\sin(2x)\)
Recall that \(\sin(2x) = 2\sin(x)\cos(x)\). Therefore, we can rewrite the denominator:
\[ 2 + \sin(2x) = 2 + 2\sin(x)\cos(x) = 2(1 + \sin(x)\cos(x)) \]
Now the integral can be expressed as:
\[ \frac{1}{\sqrt{2}} \int \frac{\cos(x) - \sin(x)}{2(1 + \sin(x)\cos(x))}dx \]

### Step 3: Split the integral
We can split the integral into two parts:
\[ \frac{1}{\sqrt{2}} \left( \int \frac{\cos(x)}{2(1 + \sin(x)\cos(x))}dx - \int \frac{\sin(x)}{2(1 + \sin(x)\cos(x))}dx \right) \]

### Step 4: Use substitution
Let \(u = \sin(x) + \cos(x)\). Then, we differentiate:
\[ du = (\cos(x) - \sin(x))dx \]
Now we can express the integrals in terms of \(u\).

### Step 5: Evaluate the integrals
The integrals can be evaluated using the substitution \(u\):
\[ \int \frac{du}{2(1 + \frac{u^2 - 1}{2})} = \int \frac{du}{u^2 + 1} \]
This integral evaluates to:
\[ \frac{1}{2} \tan^{-1}(u) + C \]

### Step 6: Substitute back
Substituting back \(u = \sin(x) + \cos(x)\):
\[ \frac{1}{\sqrt{2}} \left( \frac{1}{2} \tan^{-1}(\sin(x) + \cos(x)) + C \right) \]

### Step 7: Identify constants
Let \(A = \frac{1}{2\sqrt{2}}\) and \(B = \frac{C}{\sqrt{2}}\). Thus, we have:
\[ \int \frac{\sin\left(\frac{\pi}{4} - x\right)dx}{2 + \sin(2x)} = A \tan^{-1}(\sin(x) + \cos(x)) + B \]

### Step 8: Determine the range of \(Af(x)\)
Now, we need to find the range of \(Af(x)\), where \(f(x) = \sin(x) + \cos(x)\). The maximum and minimum values of \(\sin(x) + \cos(x)\) can be found using the identity:
\[ \sin(x) + \cos(x) = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) \]
Thus, the range of \(f(x)\) is \([- \sqrt{2}, \sqrt{2}]\).

### Step 9: Find the range of \(Af(x)\)
Multiplying by \(A = \frac{1}{2\sqrt{2}}\):
\[ \text{Range of } Af(x) = \left[-\frac{1}{2}, \frac{1}{2}\right] \]

### Final Answer
The range of \(Af(x)\) is \([- \frac{1}{2}, \frac{1}{2}]\).

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