The solution for x of the equation int_{sqrt{ | vedclass

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

Question

(D) We are given the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$.Using the standard integral formula $\int \frac{dt}{t\sqr

Vedclass · Accepted Answer

none of these

Vedclass · Answer

(D) We are given the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$.Using the standard integral formula $\int \frac{dt}{t\sqrt{t^2-1}} = \sec^{-1} t$,we evaluate the definite integral:$\left[\sec^{-1} t\right]_{\sqrt{2}}^{x} = \frac{\pi}{2}$.Substituting the limits,we get:$\sec^{-1} x - \sec^{-1} \sqrt{2} = \frac{\pi}{2}$.Since $\sec^{-1} \sqrt{2} = \frac{\pi}{4}$,the equation becomes:$\sec^{-1} x - \frac{\pi}{4} = \frac{\pi}{2}$.Solving for $\sec^{-1} x$:$\sec^{-1} x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.Therefore,$x = \sec\left(\frac{3\pi}{4}\right) = -\sqrt{2}$.Since $-\sqrt{2}$ is not among the given options,the correct choice is $(D)$.

The solution for $x$ of the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$ is

Similar Questions

Evaluate the definite integral: $\int_0^{\pi /2} \cos^2 x \, dx$

Let $\lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t - a \sin 4 t}{t^2} d t = \frac{a \sin 4 x}{x} - 1$. Find the values of $a$ and $b$.

The integral $\int_{1/4}^{3/4} \cos \left(2 \cot^{-1} \sqrt{\frac{1-x}{1+x}}\right) dx$ is equal to:

Let $f(x) = \int \frac{\sqrt{x}}{(1+x)^2} dx$ $(x \geq 0)$. Then $f(3) - f(1)$ is equal to

$\int_0^1 x \operatorname{Sin}^{-1} x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module