Question

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

• A. 1
• B. 1/2
• C. π/4
• D. None of these

Answer 1

Answer: (A)

1

Answer 2

Let I (x) = $\int_{1/e}^{\tan x}{\frac{tdt}{1 + t^{2}} +}\int_{1/e}^{\cot x}\frac{dt}{t\left( 1 + t^{2} \right)}$

= $\int_{1/e}^{\tan x}{\frac{tdt}{1 + t^{2}} +}\int_{1/e}^{\cot x}{\left( \frac{1}{t} - \frac{t}{1 + t^{2}} \right)dt}$

= $\int_{\cot x}^{\tan x}{\frac{tdt}{1 + t^{2}} +}\int_{1/e}^{\cot x}{\frac{1}{t}dt} = \left. \ \frac{1}{2}\ln(1 + t^{2}) \right|_{\cot x}^{\tan x} + \left. \ \ln t \right|_{1/e}^{\cot x}$= – ln cot x + ln cot x – ln $\frac{1}{e}$ = 1