If `int_(pi/2)^x sqrt(3-2sin^2) tdt+ int_0^y costdt=0,` then `(dy/dx)at x =pi and y=pi` is

Options

A. `sqrt3`
B. `-sqrt2`
C. `-sqrt3`
D. none of these

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given the equation:

\[ \int_{\frac{\pi}{2}}^{x} \sqrt{3 - 2\sin^2 t} \, dt + \int_{0}^{y} \cos t \, dt = 0 \]

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

Using Leibniz's rule for differentiation under the integral sign, we differentiate both integrals:

\[ \frac{d}{dx} \left( \int_{\frac{\pi}{2}}^{x} \sqrt{3 - 2\sin^2 t} \, dt \right) + \frac{d}{dx} \left( \int_{0}^{y} \cos t \, dt \right) = 0 \]

### Step 2: Apply Leibniz's rule

For the first integral, we have:

\[ \frac{d}{dx} \left( \int_{\frac{\pi}{2}}^{x} \sqrt{3 - 2\sin^2 t} \, dt \right) = \sqrt{3 - 2\sin^2 x} \]

For the second integral, we apply the chain rule:

\[ \frac{d}{dx} \left( \int_{0}^{y} \cos t \, dt \right) = \cos y \cdot \frac{dy}{dx} \]

### Step 3: Set up the equation

Combining these results, we get:

\[ \sqrt{3 - 2\sin^2 x} + \cos y \cdot \frac{dy}{dx} = 0 \]

### Step 4: Solve for \(\frac{dy}{dx}\)

Rearranging the equation gives:

\[ \cos y \cdot \frac{dy}{dx} = -\sqrt{3 - 2\sin^2 x} \]

Thus,

\[ \frac{dy}{dx} = -\frac{\sqrt{3 - 2\sin^2 x}}{\cos y} \]

### Step 5: Evaluate at \(x = \pi\) and \(y = \pi\)

Now, we need to evaluate \(\frac{dy}{dx}\) at \(x = \pi\) and \(y = \pi\):

1. Calculate \(\sqrt{3 - 2\sin^2 \pi}\):
- \(\sin \pi = 0\) so \(\sqrt{3 - 2(0)} = \sqrt{3}\)

2. Calculate \(\cos \pi\):
- \(\cos \pi = -1\)

Substituting these values into the expression for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = -\frac{\sqrt{3}}{-1} = \sqrt{3} \]

### Final Answer

Thus, the value of \(\frac{dy}{dx}\) at \(x = \pi\) and \(y = \pi\) is:

\[ \frac{dy}{dx} = \sqrt{3} \]