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 expression \( I_1 + 2I_2 \) where:

\[ I_1 = \int e^{x^2} \cos x \, dx \]
\[ I_2 = \int x e^{x^2} \sin x \, dx \]

### Step 1: Integration by Parts for \( I_1 \)

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

Then, we need to find \( du \) and \( v \):
- \( du = 2x e^{x^2} \, dx \) (using the chain rule)
- \( v = \sin x \)

Applying integration by parts:
\[ I_1 = \int u \, dv = uv - \int v \, du \]
Substituting the values:
\[ I_1 = e^{x^2} \sin x - \int \sin x \cdot 2x e^{x^2} \, dx \]
This can be rewritten as:
\[ I_1 = e^{x^2} \sin x - 2 \int x e^{x^2} \sin x \, dx \]
Notice that \( \int x e^{x^2} \sin x \, dx \) is \( I_2 \):
\[ I_1 = e^{x^2} \sin x - 2I_2 \]

### Step 2: Rearranging the Equation

Now, rearranging the equation gives us:
\[ I_1 + 2I_2 = e^{x^2} \sin x + C \]
where \( C \) is the constant of integration.

### Conclusion

Thus, we have:
\[ I_1 + 2I_2 = e^{x^2} \sin x + C \]

### Final Answer

The expression simplifies to:
\[ e^{x^2} \sin x + C \]