If int_{0}^{a} frac{dx}{1+4x^{2}} = frac{pi}{ | vedclass

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

Question

(A) Given the integral $\int_{0}^{a} \frac{dx}{1+(2x)^{2}} = \frac{\pi}{8}$.Let $2x = t$,then $2dx = dt$ or $dx = \frac{dt}{2}$.When $x = 0, t = 0$ an

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Given the integral $\int_{0}^{a} \frac{dx}{1+(2x)^{2}} = \frac{\pi}{8}$.Let $2x = t$,then $2dx = dt$ or $dx = \frac{dt}{2}$.When $x = 0, t = 0$ and when $x = a, t = 2a$.The integral becomes $\int_{0}^{2a} \frac{1}{1+t^{2}} \cdot \frac{dt}{2} = \frac{\pi}{8}$.$\frac{1}{2} [	an^{-1}(t)]_{0}^{2a} = \frac{\pi}{8}$.$	an^{-1}(2a) - 	an^{-1}(0) = \frac{\pi}{4}$.$	an^{-1}(2a) = \frac{\pi}{4}$.$2a = 	an(\frac{\pi}{4}) = 1$.$a = \frac{1}{2}$.

If $\int_{0}^{a} \frac{dx}{1+4x^{2}} = \frac{\pi}{8}$,then $a =$

Similar Questions

$\int_{0}^{\pi/4} \sqrt{1+\sin 2x} dx = \rule{1cm}{0.15mm}$

If $\theta_{1}$ and $\theta_{2}$ are respectively the smallest and the largest values of $\theta$ in $(0, 2\pi) - \{\pi\}$ which satisfy the equation $2 \cot^{2} \theta - \frac{5}{\sin \theta} + 4 = 0$,then $\int_{\theta_{1}}^{\theta_{2}} \cos^{2} 3\theta \, d\theta$ is equal to

If $\int_{1}^{2} \frac{dx}{(x^2 - 2x + 4)^{3/2}} = \frac{k}{k+5}$,then $k$ is equal to

$ \int_{0}^{2} [x^{2}] \, dx $

$\left[ \int_{0}^{2} \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}} \, dx \right]$ is equal to (where $[\cdot]$ is the $G.I.F.$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module