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

$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}\,d\theta = } $

The value of $\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ is

If $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$,then $[I] = \ldots$. Here,$[.]$ denotes the greatest integer function.

The value of $\int_{-\pi/2}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$,where $[t]$ denotes the greatest integer less than or equal to $t$,is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module