Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. None of these

Answer 1

The correct option is B $1$

Let $I={\int }_{1/e}^{\tan x}\frac{tdt}{1+t^2}+{\int }_{1/e}^{\cot x}\frac{dt}{t(1+t^2)}$
On integrating both functions, we get
$=\frac{1}{2}{\left|\log (1+t^2)\right|}_{1/e}^{\tan x}+{\left|\{\log t-\frac{1}{2}\log (1+t^2)\}\right|}_{1/e}^{\cot x}$
$=\frac{1}{2}\{\log {\sec }^{2}x-\log (1+\frac{1}{e^2})\}+\log \cot x-\log \left(\frac{1}{e}\right)-\frac{1}{2}\{\log {\text{cosec}}^{2}x-\log (1+\frac{1}{e^2})\}$
$=-\log \left(\frac{1}{e}\right)=\log e=1$