Question

The value of ${\int }_{1/e}^{\tan x}\frac{t}{1+t^2}dt+{\int }_{1/e}^{\cot x}\frac{dt}{t(1+t^2)}$ is

• A. $1/2$
• B. $1$
• C. $\pi /4$
• D. none of these

Answer 1

Answer: (B)

$1$

${\int }_{1/e}^{\tan x}\frac{t}{1+t^2}dt+{\int }_{1/e}^{\cot x}\frac{dt}{t(1+t^2)}$
$={\int }_{1/e}^{\tan x}\frac{t}{1+t^2}dt+{\int }_{1/e}^{\cot x}(\frac{1}{t}-\frac{1}{1+t^2})dt$
$=\frac{1}{2}{\left[\log (1+t^2)\right]}_{1/e}^{\tan x}+{\left[\log t\right]}_{1/e}^{\cot x}-{\left[{\tan }^{-1}t\right]}_{1/e}^{\cot x}=1$