If y = sqrt{sec x + sqrt{sec x + sqrt{sec x + | vedclass

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(B) Given $y = \sqrt{\sec x + y}$.Squaring both sides,we get $y^2 = \sec x + y$.Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx

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$1$

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(B) Given $y = \sqrt{\sec x + y}$.Squaring both sides,we get $y^2 = \sec x + y$.Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = \sec x 	an x + \frac{dy}{dx}$.Rearranging the terms,we have $(2y - 1) \frac{dy}{dx} = \sec x 	an x$.Now,we need to evaluate the integral $I = \int_{0}^{\pi/3} (2y - 1) \frac{dy}{dx} \, dx$.Substituting the expression derived above,$I = \int_{0}^{\pi/3} \sec x 	an x \, dx$.The integral of $\sec x 	an x$ is $\sec x$.Thus,$I = [\sec x]_{0}^{\pi/3} = \sec(\pi/3) - \sec(0) = 2 - 1 = 1$.

If $y = \sqrt{\sec x + \sqrt{\sec x + \sqrt{\sec x + \dots \infty}}} \,,$ then the value of $\int_{0}^{\pi/3} (2y - 1) \frac{dy}{dx} \, dx$ is equal to $(\sec x > 0)$ -

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