Question

The value of ${\int }_{1/e}^{\tan x}\frac{t}{1+t^2}dt+{\int }_{1/e}^{\cot x}\frac{1}{t(1+t^2)}dt$, where $x\in (\pi /6,\pi /3)$, is equal to?

• A. $0$
• B. $2$
• C. $1$
• D. Cannot

Answer 1

The correct option is C $1$

$\begin{array}{c}{\int }_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+{\int }_{\frac{1}{e}}^{\cot x}\frac{1}{t\left(1+t^2\right)}dt\\ t=\frac{1}{u}\\ \frac{dt}{du}=-\frac{1}{u^2}\\ {\int }_e^{\tan x}\frac{u}{1+\frac{1}{u^2}}-\frac{du}{u^2}\\ {\int }_{\frac{1}{e}}^{\tan x}\frac{tdt}{1+t^2}+-{\int }_e^{\tan x}\frac{u}{1+u^2}du\\ {\int }_{\frac{1}{e}}^{\tan x}\frac{tdt}{1+t^2}+{\int }_{\tan x}^e\frac{t}{1+t^2}dt\end{array}$

$\begin{array}{c}{\int }_{\frac{1}{e}}^e\frac{t}{1+t^2}dt=\frac{1}{2}{\left[\ln \left(1+t^2\right)\right]}_{\frac{1}{e}}^e\\ \frac{1}{2}\left[\ln \left(1+e^2\right)-\ln \left(1+\frac{1}{e^2}\right)\right]\\ \frac{1}{2}\left[\ln \left(1+e^2\right)-\ln \left(1+e^2\right)+2\right]\\ =1\end{array}$