If `int_(pi//3)^(x) sqrt(3-2sin^(2)u) du + int_(0)^(y) cos t dt= 0`, then `(dy)/(dx)` is equal to

Options

A. `(sqrt""(4-3sin^(2)x))/(cos y)`
B. `(-sqrt""(3-2 sin^(2)x))/(cos y)`
C. `sqrt""(3-2sin^(2)x) + cos y`
D. None of these

Answer

To solve the problem, we start with the given equation:

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

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

Using the Fundamental Theorem of Calculus and the chain rule, we differentiate the left-hand side:

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

### Step 2: Apply the Fundamental Theorem of Calculus

For the first integral:

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

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

Now we can set up the equation from the differentiated terms:

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

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

Rearranging the equation gives:

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

Now, we can isolate \(\frac{dy}{dx}\):

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

### Final Result

Thus, we find that:

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

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