Let f be a function defined on the interval `[0,2pi]` such that `int_(0)^(x)(f^(')(t)-sin2t)dt=int_(x)^(0)f(t)tantdt` and `f(0)=1`. Then the maximum value of `f(x)`is…………………..

Explanation

To find the maximum value of the function \( f(x) \) defined by the equation

\[ \int_{0}^{x} (f'(t) - \sin 2t) dt = \int_{x}^{0} f(t) \tan t dt \]

with the condition \( f(0) = 1 \), we will follow these steps:

### Step 1: Differentiate both sides

We start by differentiating both sides of the equation with respect to \( x \):

\[ f'(x) - \sin(2x) = -f(x) \tan(x) \]

### Step 2: Rearranging the equation

Rearranging the equation gives us:

\[ f'(x) + f(x) \tan(x) = \sin(2x) \]

### Step 3: Recognizing the form of the equation

This is a first-order linear differential equation of the form:

\[ y' + P(x)y = Q(x) \]

where \( y = f(x) \), \( P(x) = \tan(x) \), and \( Q(x) = \sin(2x) \).

### Step 4: Finding the integrating factor

The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) dx} = e^{\int \tan(x) dx} = e^{-\ln(\cos(x))} = \sec(x) \]

### Step 5: Multiplying through by the integrating factor

Multiplying the entire differential equation by \( \sec(x) \):

\[ \sec(x) f'(x) + \sec(x) f(x) \tan(x) = \sec(x) \sin(2x) \]

This simplifies to:

\[ \frac{d}{dx} (f(x) \sec(x)) = \sec(x) \sin(2x) \]

### Step 6: Integrating both sides

Integrating both sides gives:

\[ f(x) \sec(x) = \int \sec(x) \sin(2x) dx + C \]

### Step 7: Finding the integral

Using the identity \( \sin(2x) = 2 \sin(x) \cos(x) \):

\[ \int \sec(x) \sin(2x) dx = \int 2 \sin(x) dx = -2 \cos(x) + C \]

So we have:

\[ f(x) \sec(x) = -2 \cos(x) + C \]

### Step 8: Solving for \( f(x) \)

Multiplying through by \( \cos(x) \):

\[ f(x) = -2 \cos^2(x) + C \cos(x) \]

### Step 9: Using the initial condition

Using the condition \( f(0) = 1 \):

\[ f(0) = -2 \cdot 1 + C \cdot 1 = 1 \implies C - 2 = 1 \implies C = 3 \]

Thus, we have:

\[ f(x) = -2 \cos^2(x) + 3 \cos(x) \]

### Step 10: Finding the maximum value

To find the maximum value of \( f(x) \), we need to differentiate \( f(x) \):

\[ f'(x) = -4 \cos(x) \sin(x) - 3 \sin(x) = \sin(x)(-4 \cos(x) - 3) \]

Setting \( f'(x) = 0 \):

\[ \sin(x)(-4 \cos(x) - 3) = 0 \]

This gives us two cases:

1. \( \sin(x) = 0 \) which gives \( x = 0, \pi, 2\pi \)
2. \( -4 \cos(x) - 3 = 0 \) which gives \( \cos(x) = -\frac{3}{4} \)

### Step 11: Evaluating \( f(x) \) at critical points

For \( x = 0 \):

\[ f(0) = 1 \]

For \( x = \pi \):

\[ f(\pi) = -2(-1)^2 + 3(-1) = -2 - 3 = -5 \]

For \( x = 2\pi \):

\[ f(2\pi) = -2(1)^2 + 3(1) = -2 + 3 = 1 \]

For \( \cos(x) = -\frac{3}{4} \):

Calculating \( f(x) \) at \( \cos(x) = -\frac{3}{4} \):

\[ f(x) = -2 \left(-\frac{3}{4}\right)^2 + 3\left(-\frac{3}{4}\right) = -2 \cdot \frac{9}{16} - \frac{9}{4} = -\frac{18}{16} - \frac{36}{16} = -\frac{54}{16} = -\frac{27}{8} \]

### Conclusion

The maximum value of \( f(x) \) occurs at \( x = 0 \) or \( x = 2\pi \):

\[ \text{Maximum value of } f(x) = 1 \]