If `int_(sin x)^(1) t^(2) f(t) dt =1- sin x, x in (0, (pi)/(2))` then `f((1)/(sqrt3))` equal to

Options

A. 3
B. `1//3`
C. `1//sqrt3`
D. `sqrt3`

Answer

To solve the problem, we need to find the value of \( f\left(\frac{1}{\sqrt{3}}\right) \) given the equation:

\[ \int_{\sin x}^{1} t^{2} f(t) \, dt = 1 - \sin x, \quad x \in \left(0, \frac{\pi}{2}\right) \]

### Step 1: Differentiate both sides with respect to \( x \)

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

\[ \frac{d}{dx} \left( \int_{\sin x}^{1} t^{2} f(t) \, dt \right) = \frac{d}{dx} (1 - \sin x) \]

### Step 2: Apply Leibniz's rule for differentiation under the integral sign

Using Leibniz's rule for differentiation under the integral sign, we have:

\[ \frac{d}{dx} \left( \int_{\sin x}^{1} t^{2} f(t) \, dt \right) = t^{2} f(t) \bigg|_{t=1} \cdot \frac{d}{dx}(1) - t^{2} f(t) \bigg|_{t=\sin x} \cdot \frac{d}{dx}(\sin x) \]

This simplifies to:

\[ 0 - \sin^2 x \cdot f(\sin x) \cdot \cos x \]

### Step 3: Differentiate the right-hand side

Differentiating the right-hand side gives:

\[ -\cos x \]

### Step 4: Set the derivatives equal to each other

Setting the two results equal gives us:

\[ -\sin^2 x \cdot f(\sin x) \cdot \cos x = -\cos x \]

### Step 5: Simplify the equation

If we cancel out \(-\cos x\) (noting that \(\cos x \neq 0\) in the interval \( (0, \frac{\pi}{2})\)), we get:

\[ \sin^2 x \cdot f(\sin x) = 1 \]

### Step 6: Solve for \( f(\sin x) \)

From the equation above, we can express \( f(\sin x) \):

\[ f(\sin x) = \frac{1}{\sin^2 x} \]

### Step 7: Substitute \( \sin x = \frac{1}{\sqrt{3}} \)

Now, we need to find \( f\left(\frac{1}{\sqrt{3}}\right) \). We substitute \( \sin x = \frac{1}{\sqrt{3}} \):

\[ f\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{\left(\frac{1}{\sqrt{3}}\right)^2} = \frac{1}{\frac{1}{3}} = 3 \]

### Conclusion

Thus, the value of \( f\left(\frac{1}{\sqrt{3}}\right) \) is:

\[ \boxed{3} \]