If int_{0}^{1} tan ^{-1} x , dx = p,then the | vedclass

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

Question

(B) Let $I = \int_{0}^{1} 	an ^{-1}\left(\frac{1-x}{1+x}ight) \, dx$. Using the trigonometric identity $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1}\le

Vedclass · Accepted Answer

$\frac{\pi}{4} - p$

Vedclass · Answer

(B) Let $I = \int_{0}^{1} 	an ^{-1}\left(\frac{1-x}{1+x}ight) \, dx$. Using the trigonometric identity $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$,we can write $	an ^{-1}\left(\frac{1-x}{1+x}ight) = 	an ^{-1}(1) - 	an ^{-1}(x)$. Since $	an ^{-1}(1) = \frac{\pi}{4}$,the integral becomes: $I = \int_{0}^{1} \left( \frac{\pi}{4} - 	an ^{-1}(x) ight) \, dx$. $I = \int_{0}^{1} \frac{\pi}{4} \, dx - \int_{0}^{1} 	an ^{-1}(x) \, dx$. $I = \left[ \frac{\pi}{4} x ight]_{0}^{1} - p$. $I = \frac{\pi}{4}(1 - 0) - p$. $I = \frac{\pi}{4} - p$.

If $\int_{0}^{1} \tan ^{-1} x \, dx = p$,then the value of $\int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) \, dx$ is

Similar Questions

$A$ function $f(x)$ satisfies $f(x) = f(\frac{c}{x})$ for some real number $c$ $(c > 1)$ and $\forall\, x > 0$. If $\int_{1}^{\sqrt{c}} \frac{f(x)}{x} dx = 3$,then the value of $\int_{1}^{c} \frac{f(x)}{x} dx$ is

$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ is equal to

$\int_0^{2a} \frac{f(x)}{f(x) + f(2a - x)} \, dx = $

If $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$, then the value of $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ is

$\int_{-1}^1 \left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module