The value of `int_(1//e)^(tan x) (t)/(1+ t^(2)) dt+ int_(1//e)^(cot x) (1)/(t(1+ t^(2)))dt` is

Options

A. `-1`
B. 1
C. 0
D. None of these

Answer

To solve the given problem, we need to evaluate the integral:

\[ I = \int_{\frac{1}{e}}^{\tan x} \frac{t}{1 + t^2} \, dt + \int_{\frac{1}{e}}^{\cot x} \frac{1}{t(1 + t^2)} \, dt \]

Let's denote the two integrals as \( I_1 \) and \( I_2 \):

1. **Evaluate \( I_1 \)**:
\[ I_1 = \int_{\frac{1}{e}}^{\tan x} \frac{t}{1 + t^2} \, dt \]
To solve this integral, we can use the substitution \( \theta = \tan^{-1}(t) \). Then, we have:
\[ d\theta = \frac{1}{1 + t^2} dt \quad \Rightarrow \quad dt = (1 + t^2) d\theta \]
Thus, we can rewrite the integral as:
\[ I_1 = \int_{\tan^{-1}(\frac{1}{e})}^{\tan^{-1}(\tan x)} t \, d\theta \]
Since \( t = \tan \theta \), we have:
\[ I_1 = \int_{\tan^{-1}(\frac{1}{e})}^{x} \tan \theta \, d\theta \]
The integral of \( \tan \theta \) is:
\[ \int \tan \theta \, d\theta = -\log(\cos \theta) \]
Therefore, we can evaluate:
\[ I_1 = -\log(\cos x) + \log(\cos(\tan^{-1}(\frac{1}{e}))) \]
We know that \( \cos(\tan^{-1}(t)) = \frac{1}{\sqrt{1 + t^2}} \), so:
\[ \cos(\tan^{-1}(\frac{1}{e})) = \frac{1}{\sqrt{1 + \frac{1}{e^2}}} = \frac{e}{\sqrt{e^2 + 1}} \]
Thus:
\[ I_1 = -\log(\cos x) + \log\left(\frac{e}{\sqrt{e^2 + 1}}\right) \]

2. **Evaluate \( I_2 \)**:
\[ I_2 = \int_{\frac{1}{e}}^{\cot x} \frac{1}{t(1 + t^2)} \, dt \]
We can use partial fractions to simplify:
\[ \frac{1}{t(1 + t^2)} = \frac{A}{t} + \frac{Bt + C}{1 + t^2} \]
Solving for \( A, B, C \) gives \( A = 1 \), \( B = -1 \), and \( C = 0 \). Thus:
\[ I_2 = \int_{\frac{1}{e}}^{\cot x} \left(\frac{1}{t} - \frac{t}{1 + t^2}\right) dt \]
This can be separated into two integrals:
\[ I_2 = \int_{\frac{1}{e}}^{\cot x} \frac{1}{t} dt - \int_{\frac{1}{e}}^{\cot x} \frac{t}{1 + t^2} dt \]
The first integral evaluates to:
\[ \log(\cot x) - \log\left(\frac{1}{e}\right) = \log(\cot x) + 1 \]
The second integral is the same as \( I_1 \) but with different limits, giving:
\[ I_2 = \log(\cot x) + 1 - I_1 \]

3. **Combine \( I_1 \) and \( I_2 \)**:
\[ I = I_1 + I_2 = I_1 + \left(\log(\cot x) + 1 - I_1\right) \]
Thus:
\[ I = \log(\cot x) + 1 \]

Finally, we conclude that the value of the integral is:

\[ \boxed{1} \]