What is `inte^(ln(tanx))` dx equal to ? Where e is the constant of integration.

Options

A. `ln|tanx|+c`
B. `ln|secx|+c`
C. `tanx+c`
D. `e^(tanx)+c`

Answer

To solve the integral \( \int e^{\ln(\tan x)} \, dx \), we can simplify the expression inside the integral first.

### Step-by-step Solution:

1. **Simplify the Integrand**:
\[ e^{\ln(\tan x)} = \tan x \]
This is because the exponential function and the natural logarithm are inverse functions.

2. **Set Up the Integral**:
Now we can rewrite the integral:
\[ \int e^{\ln(\tan x)} \, dx = \int \tan x \, dx \]

3. **Integrate \( \tan x \)**:
The integral of \( \tan x \) can be computed using the identity:
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus, we can use substitution. Let:
\[ u = \cos x \quad \Rightarrow \quad du = -\sin x \, dx \quad \Rightarrow \quad -du = \sin x \, dx \]
Therefore, the integral becomes:
\[ \int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx = \int -\frac{1}{u} \, du \]

4. **Compute the Integral**:
Now we can integrate:
\[ -\int \frac{1}{u} \, du = -\ln |u| + C = -\ln |\cos x| + C \]

5. **Simplify the Result**:
Using the property of logarithms, we can rewrite the result:
\[ -\ln |\cos x| = \ln \left( \frac{1}{|\cos x|} \right) = \ln(\sec x) \]
Therefore, the final result is:
\[ \int e^{\ln(\tan x)} \, dx = \ln(\sec x) + C \]

### Final Answer:
\[ \int e^{\ln(\tan x)} \, dx = \ln(\sec x) + C \]