The value of the integral int limits_{1 / 2}^ | vedclass

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

Question

(D) Let $I = \int \limits_{1 / 2}^2 \frac{	an ^{-1} x}{x} dx$  ... $(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(\frac{ab}{x}) dx$,we substi

Vedclass · Accepted Answer

$\frac{\pi}{2} \log _{ e } 2$

Vedclass · Answer

(D) Let $I = \int \limits_{1 / 2}^2 \frac{	an ^{-1} x}{x} dx$  ... $(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(\frac{ab}{x}) dx$,we substitute $x = \frac{1}{t}$,so $dx = -\frac{1}{t^2} dt$.When $x = 1/2, t = 2$ and when $x = 2, t = 1/2$.$I = \int \limits_2^{1 / 2} \frac{	an ^{-1}(1/t)}{1/t} \cdot (-\frac{1}{t^2}) dt = \int \limits_{1 / 2}^2 \frac{	an ^{-1}(1/t)}{t} dt$.Since $	an^{-1}(1/t) = \cot^{-1} t$ for $t > 0$,we have $I = \int \limits_{1 / 2}^2 \frac{\cot ^{-1} t}{t} dt = \int \limits_{1 / 2}^2 \frac{\cot ^{-1} x}{x} dx$ ... (ii)Adding $(i)$ and (ii):$2I = \int \limits_{1 / 2}^2 \frac{	an ^{-1} x + \cot ^{-1} x}{x} dx = \int \limits_{1 / 2}^2 \frac{\pi / 2}{x} dx$.$2I = \frac{\pi}{2} [\ln x]_{1/2}^2 = \frac{\pi}{2} (\ln 2 - \ln(1/2)) = \frac{\pi}{2} (\ln 2 - (-\ln 2)) = \frac{\pi}{2} (2 \ln 2) = \pi \ln 2$.Therefore,$I = \frac{\pi}{2} \ln 2$.

The value of the integral $\int \limits_{1 / 2}^2 \frac{\tan ^{-1} x}{x} d x$ is equal to

Similar Questions

$\int_{-1}^3 \left(\tan^{-1}\left(\frac{x}{x^2+1}\right) + \tan^{-1}\left(\frac{x^2+1}{x}\right)\right) dx =$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^2 + \log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x\right) dx =$

If $f(x) = \int\limits_0^\pi {\frac{t \sin t \, dt}{\sqrt{1 + \tan^2 x \sin^2 t}}}$ for $0 < x < \frac{\pi}{2}$,then which of the following is true?

The value of $\int_{0}^{4042} \frac{\sqrt{x} \, dx}{\sqrt{x}+\sqrt{4042-x}}$ is equal to

$\int_0^{\frac{\pi}{2}} \frac{x \tan x \sec^2 x}{\tan^4 x + 1} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module