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

$\int_{a-6}^{b-6} f(x+6) dx$ is equal to

$\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}\right] d x=$

Evaluate $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^{2} x \, dx$

If $n$ is a positive integer and $[x]$ is the greatest integer not exceeding $x$,then $\int_0^n {\{x - [x]\} \,dx}$ equals

For $f(x) = x^4 + |x|$,let $I_1 = \int_{0}^{\pi} f(\cos x) dx$ and $I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$. Then $\frac{I_1}{I_2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module