he value of the integral `inte^(sin^(2x))(cosx+cos^3x)sinx dx` is

Options

A. `1/2 e^(sin^(2)x)(3-sin^(2)x)+C`
B. `e^( sin^(2)x)(1+1/2 cos^(2)x)+C`
C. `e^(sin^(2)x)(3 cos^(2)x + 2 sin^(2)x)+C`
D. `e^(sin^(2)x)( 2 cos^(2)x+3 sin^(2)x)+C`

Answer

To solve the integral \( \int e^{\sin^2 x} (\cos x + \cos^3 x) \sin x \, dx \), we will follow these steps:

### Step 1: Simplify the Integral
We start with the integral:
\[ I = \int e^{\sin^2 x} (\cos x + \cos^3 x) \sin x \, dx \]
We can factor out \( \cos x \) from the expression:
\[ I = \int e^{\sin^2 x} \cos x (1 + \cos^2 x) \sin x \, dx \]

### Step 2: Use Substitution
Let \( t = \sin^2 x \). Then, the derivative \( dt \) is given by:
\[ dt = 2 \sin x \cos x \, dx \quad \Rightarrow \quad \frac{dt}{2} = \sin x \cos x \, dx \]
Thus, we can rewrite the integral in terms of \( t \):
\[ I = \int e^t \cos x (1 + \cos^2 x) \frac{dt}{2} \]

### Step 3: Express \( \cos x \) in Terms of \( t \)
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \cos^2 x = 1 - t \quad \Rightarrow \quad \cos x = \sqrt{1 - t} \]
Now substituting this back into the integral:
\[ I = \frac{1}{2} \int e^t \sqrt{1 - t} (1 + (1 - t)) \, dt \]
This simplifies to:
\[ I = \frac{1}{2} \int e^t \sqrt{1 - t} (2 - t) \, dt \]

### Step 4: Split the Integral
Now we can split the integral into two parts:
\[ I = \frac{1}{2} \left( 2 \int e^t \sqrt{1 - t} \, dt - \int t e^t \sqrt{1 - t} \, dt \right) \]

### Step 5: Integrate by Parts
For the second integral, we can use integration by parts:
Let \( u = \sqrt{1 - t} \) and \( dv = t e^t dt \). Then, we differentiate and integrate accordingly:
\[ du = -\frac{1}{2\sqrt{1 - t}} dt, \quad v = e^t(t - 1) \]
Using integration by parts, we can evaluate the integral.

### Step 6: Combine and Simplify
After evaluating both integrals, we combine the results and factor out \( e^{\sin^2 x} \) to express the final answer in terms of \( \sin^2 x \).

### Final Result
After simplification, we find:
\[ I = \frac{1}{2} e^{\sin^2 x} (3 - \sin^2 x) + C \]
Thus, the value of the integral is:
\[ \frac{1}{2} e^{\sin^2 x} (3 - \sin^2 x) + C \]