Evaluate `int sin 4x.e^(tan^(2)x)dx`.

Answer

To evaluate the integral \( I = \int \sin(4x) e^{\tan^2 x} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Sine Function**:
We can express \( \sin(4x) \) using the double angle formula:
\[ \sin(4x) = \sin(2 \cdot 2x) = 2 \sin(2x) \cos(2x) \]
Thus, we can rewrite the integral:
\[ I = \int 2 \sin(2x) \cos(2x) e^{\tan^2 x} \, dx \]

2. **Substituting for Sine and Cosine**:
Using the identities for sine and cosine in terms of tangent:
\[ \sin(2x) = \frac{2 \tan x}{1 + \tan^2 x}, \quad \cos(2x) = \frac{1 - \tan^2 x}{1 + \tan^2 x} \]
We substitute these into the integral:
\[ I = 2 \int \left( \frac{2 \tan x}{1 + \tan^2 x} \right) \left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) e^{\tan^2 x} \, dx \]

3. **Simplifying the Integral**:
The integral simplifies to:
\[ I = 2 \int \frac{2 \tan x (1 - \tan^2 x) e^{\tan^2 x}}{(1 + \tan^2 x)^2} \, dx \]

4. **Substitution**:
Let \( t = \tan^2 x \). Then, differentiating gives:
\[ dt = 2 \tan x \sec^2 x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2 \tan x \sec^2 x} \]
Substituting this into the integral gives:
\[ I = 2 \int \frac{2 \tan x (1 - t) e^t}{(1 + t)^2} \cdot \frac{dt}{2 \tan x \sec^2 x} \]
This simplifies to:
\[ I = \int \frac{(1 - t) e^t}{(1 + t)^2} \, dt \]

5. **Splitting the Integral**:
We can split the integral:
\[ I = \int \frac{e^t}{(1 + t)^2} \, dt - \int \frac{t e^t}{(1 + t)^2} \, dt \]

6. **Integration by Parts**:
The second integral can be solved using integration by parts. Let:
\[ u = \frac{1}{1 + t}, \quad dv = e^t \, dt \quad \Rightarrow \quad du = -\frac{1}{(1+t)^2} \, dt, \quad v = e^t \]
Thus:
\[ \int \frac{t e^t}{(1 + t)^2} \, dt = \frac{t e^t}{1+t} - \int \frac{e^t}{1+t} \, dt \]

7. **Final Expression**:
After evaluating the integrals, we substitute back \( t = \tan^2 x \) to get the final result:
\[ I = -2 e^{\tan^2 x} \cdot \frac{1}{\sec^2 x} + C = -2 e^{\tan^2 x} \cos^2 x + C \]

### Final Answer:
\[ I = -2 e^{\tan^2 x} \cos^2 x + C \]