Question

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

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

Let $f\left(x\right)=\underset{1/e}{\overset{\tan x}{\int }}\frac{tdt}{1+t^2}+\underset{1/e}{\overset{\cot x}{\int }}\frac{dt}{t(1+t^2)}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\tan x}{1+{\tan }^{2}x}{\sec }^{2}x+\frac{1}{\cot x(1+{\cot }^{2}x)}(-{\text{cosec}}^{2}x)$
$\Rightarrow f^{\prime }\left(x\right)=\tan x-\tan x=0$
$\Rightarrow f\left(x\right)$ is a constant function.
Put $x=\frac{\pi }{4}$
$\therefore f\left(\frac{\pi }{4}\right)=\underset{\frac{1}{e}}{\overset{1}{\int }}\frac{tdt}{(1+t^2)}+\underset{\frac{1}{e}}{\overset{1}{\int }}\frac{dt}{t(1+t^2)}$
$\Rightarrow f\left(x\right)=\underset{\frac{1}{e}}{\overset{1}{\int }}\frac{1}{t}dt=[\ln |t|{]}_{1/e}^1=1$