The value of x that satisfies the equation in | vedclass

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

Question

(D) We are given the integral equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$.Recall the standard integral formula $\int \fr

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We are given the integral equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$.Recall the standard integral formula $\int \frac{dt}{|t| \sqrt{t^2-1}} = \sec^{-1}(t) + C$.Applying the limits of integration,we get:$\left[ \sec^{-1}(t) \right]_{\sqrt{2}}^x = \frac{\pi}{12}$.Substituting the limits,we have $\sec^{-1}(x) - \sec^{-1}(\sqrt{2}) = \frac{\pi}{12}$.Since $\sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$,the equation becomes $\sec^{-1}(x) - \frac{\pi}{4} = \frac{\pi}{12}$.Adding $\frac{\pi}{4}$ to both sides,we get $\sec^{-1}(x) = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi + 3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.Therefore,$x = \sec\left(\frac{\pi}{3}\right) = 2$.

The value of $x$ that satisfies the equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$ is

Similar Questions

$\int_{-3}^3 |2-x| dx =$

If $f : R \to R$ is a continuous function such that $f(x) = \int\limits_1^x {tf(t)dt}$,then the correct statement is -

$\int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx =$

$\int_{-2}^{1} [x+1] \, dx =$ (Where $[x]$ is the greatest integer function not greater than $x$)

If $\int_0^{\frac{\pi}{3}} \cos^4 x \, dx = a\pi + b\sqrt{3}$,where $a$ and $b$ are rational numbers,then $9a + 8b$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module