If `phi(x)=(1)/(sqrt(x))int_(pi/4)^(x)(4sqrt(2)sin t-3 phi'(t))dt,x>0,` then `phi'((pi)/(4))` is equal to

Options

A. `(4)/(6+sqrt(pi))`
B. `(8)/(6+sqrt(pi))`
C. `(8)/(sqrt(pi))`
D. `(4)/(6-sqrt(pi))`

Answer

To solve the problem, we need to find the value of \( \phi'(\frac{\pi}{4}) \) given the function \( \phi(x) \) defined as:

\[ \phi(x) = \frac{1}{\sqrt{x}} \int_{\frac{\pi}{4}}^{x} \left( 4\sqrt{2} \sin t - 3 \phi'(t) \right) dt, \quad x > 0 \]

### Step 1: Evaluate \( \phi(\frac{\pi}{4}) \)

First, we substitute \( x = \frac{\pi}{4} \) into the equation for \( \phi(x) \):

\[ \phi\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\frac{\pi}{4}}} \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \left( 4\sqrt{2} \sin t - 3 \phi'(t) \right) dt \]

Since the upper and lower limits of the integral are the same, the integral evaluates to 0:

\[ \phi\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\frac{\pi}{4}}} \cdot 0 = 0 \]

### Step 2: Differentiate \( \phi(x) \)

Next, we differentiate \( \phi(x) \) with respect to \( x \) using the product rule and the Fundamental Theorem of Calculus:

\[ \phi'(x) = \frac{d}{dx} \left( \frac{1}{\sqrt{x}} \right) \int_{\frac{\pi}{4}}^{x} \left( 4\sqrt{2} \sin t - 3 \phi'(t) \right) dt + \frac{1}{\sqrt{x}} \left( 4\sqrt{2} \sin x - 3 \phi'(x) \right) \]

Using the derivative of \( \frac{1}{\sqrt{x}} \):

\[ \frac{d}{dx} \left( \frac{1}{\sqrt{x}} \right) = -\frac{1}{2} x^{-\frac{3}{2}} \]

Thus, we have:

\[ \phi'(x) = -\frac{1}{2} x^{-\frac{3}{2}} \int_{\frac{\pi}{4}}^{x} \left( 4\sqrt{2} \sin t - 3 \phi'(t) \right) dt + \frac{1}{\sqrt{x}} \left( 4\sqrt{2} \sin x - 3 \phi'(x) \right) \]

### Step 3: Substitute \( x = \frac{\pi}{4} \)

Now we substitute \( x = \frac{\pi}{4} \) into the differentiated equation:

\[ \phi'\left(\frac{\pi}{4}\right) = -\frac{1}{2} \left(\frac{\pi}{4}\right)^{-\frac{3}{2}} \cdot 0 + \frac{1}{\sqrt{\frac{\pi}{4}}} \left( 4\sqrt{2} \sin\left(\frac{\pi}{4}\right) - 3 \phi'\left(\frac{\pi}{4}\right) \right) \]

Since \( \phi\left(\frac{\pi}{4}\right) = 0 \), the first term is 0. Now we simplify the second term:

\[ \phi'\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\frac{\pi}{4}}} \left( 4\sqrt{2} \cdot \frac{1}{\sqrt{2}} - 3 \phi'\left(\frac{\pi}{4}\right) \right) \]

This simplifies to:

\[ \phi'\left(\frac{\pi}{4}\right) = \frac{2}{\sqrt{\pi}} \left( 4 - 3 \phi'\left(\frac{\pi}{4}\right) \right) \]

### Step 4: Solve for \( \phi'(\frac{\pi}{4}) \)

Rearranging gives:

\[ \phi'\left(\frac{\pi}{4}\right) + \frac{6}{\sqrt{\pi}} \phi'\left(\frac{\pi}{4}\right) = \frac{8}{\sqrt{\pi}} \]

Combining terms:

\[ \left(1 + \frac{6}{\sqrt{\pi}}\right) \phi'\left(\frac{\pi}{4}\right) = \frac{8}{\sqrt{\pi}} \]

Solving for \( \phi'\left(\frac{\pi}{4}\right) \):

\[ \phi'\left(\frac{\pi}{4}\right) = \frac{8/\sqrt{\pi}}{1 + \frac{6}{\sqrt{\pi}}} = \frac{8}{\sqrt{\pi} + 6} \]

### Final Answer

Thus, the value of \( \phi'\left(\frac{\pi}{4}\right) \) is:

\[ \phi'\left(\frac{\pi}{4}\right) = \frac{8}{\sqrt{\pi}} \]