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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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 $(\sec x > 0)$ -

Similar Questions

Evaluate the integral: $\int_0^{\pi /2} \frac{1 + 2\cos x}{(2 + \cos x)^2} dx$

$\int_0^\infty {{e^{ - 2x}}(\sin 2x + \cos 2x)\,dx = } $

Let $f(x) = \begin{cases} \frac{x}{\sin x}, & x \in (0, 1) \\ 1, & x = 0 \end{cases}$. Consider the integral $I_n = \sqrt{n} \int_0^{1/n} f(x) e^{-nx} dx$. Then,$\lim_{n \to \infty} I_n$ is:

$\int_0^{\pi / 4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ is equal to

Evaluate $\int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module