The value of the integral _1/2^2^-1xx,dx is equal to:

Question

The value of the integral $\displaystyle\int_{1/2}^{2}\frac{	an^{-1}x}{x}\,dx$ is equal to:

Accepted Answer

$$I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ ..... (i)

$$x \to {1 \over x}$$

$$I = \int\limits_{{1 \over 2}}^2 {{1 \over x}{{\tan }^{ - 1}}{1 \over x}dx} $$ ..... (ii)

$$2I = \int\limits_{{1 \over 2}}^2 {{1 \over x}\,.\,{\pi  \over 2}dx} $$

$$ = \left. {{\pi  \over 2}\ln x} \right|_{{1 \over 2}}^2 = \pi \ln 2$$

$$ \Rightarrow I = {\pi  \over 2}\ln 2$$

Solutions · 1