Consider the integrals `I_(1)=inte^(x^(2))cosxdx` and `I_(2)=intxe^(x^(2))sinxdx`. Then `I_(1)+2I_(2)` simplifies to (Where, c is the constant of integration)

Options

A. `e^(x)sin x+c`
B. `e^(x^(2))cos x+c`
C. `2e^(x^(2))sinx+c`
D. `e^(x^(2))sin x+c`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and find the expression for \( I_1 + 2I_2 \).

### Step 1: Define the integrals
Let:
\[ I_1 = \int e^{x^2} \cos x \, dx \]
\[ I_2 = \int x e^{x^2} \sin x \, dx \]

### Step 2: Integrate \( I_1 \) by parts
We will use integration by parts on \( I_1 \). Let:
- \( u = \cos x \) (first function)
- \( dv = e^{x^2} \, dx \) (second function)

Then we need to find \( du \) and \( v \):
\[ du = -\sin x \, dx \]
To find \( v \), we note that \( \int e^{x^2} \, dx \) does not have a simple antiderivative, but we can express it as:
\[ v = \int e^{x^2} \, dx \]

Using integration by parts:
\[ I_1 = uv - \int v \, du \]
Substituting the values:
\[ I_1 = \cos x \cdot \int e^{x^2} \, dx + \int \sin x \cdot \int e^{x^2} \, dx \, dx \]

### Step 3: Integrate \( I_2 \) by parts
Now, we will integrate \( I_2 \) by parts as well. Let:
- \( u = \sin x \) (first function)
- \( dv = x e^{x^2} \, dx \) (second function)

Then we need to find \( du \) and \( v \):
\[ du = \cos x \, dx \]
To find \( v \), we can integrate \( dv \):
\[ v = \frac{1}{2} e^{x^2} \]

Using integration by parts:
\[ I_2 = uv - \int v \, du \]
Substituting the values:
\[ I_2 = \sin x \cdot \frac{1}{2} e^{x^2} - \int \frac{1}{2} e^{x^2} \cos x \, dx \]
This gives us:
\[ I_2 = \frac{1}{2} e^{x^2} \sin x - \frac{1}{2} I_1 \]

### Step 4: Combine \( I_1 \) and \( I_2 \)
Now, we can express \( I_1 + 2I_2 \):
\[ I_1 + 2I_2 = I_1 + 2\left(\frac{1}{2} e^{x^2} \sin x - \frac{1}{2} I_1\right) \]
This simplifies to:
\[ I_1 + e^{x^2} \sin x - I_1 = e^{x^2} \sin x \]

### Step 5: Final expression
Thus, we have:
\[ I_1 + 2I_2 = e^{x^2} \sin x + C \]
where \( C \) is the constant of integration.

### Conclusion
The final answer is:
\[ I_1 + 2I_2 = e^{x^2} \sin x + C \]