If `inte^(sintheta)(sintheta+sec^2theta)"d"theta` is equal to `f(theta)+C` (where , C is the constant of integration) and f(0) = 0 , then the value of `f(pi/4)` is

Options

A. `e^(sqrt(2))`
B. `e^(1/sqrt(2))`
C. `e^2`
D. `e^(1/2)`

Answer

To solve the integral \( \int e^{\sin \theta} (\sin \theta + \sec^2 \theta) \, d\theta \) and find the value of \( f(\frac{\pi}{4}) \) given that \( f(0) = 0 \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int e^{\sin \theta} (\sin \theta + \sec^2 \theta) \, d\theta \]
This can be split into two parts:
\[ \int e^{\sin \theta} \sin \theta \, d\theta + \int e^{\sin \theta} \sec^2 \theta \, d\theta \]

### Step 2: Integration by Parts
For the first integral \( \int e^{\sin \theta} \sin \theta \, d\theta \), we will use integration by parts. Let:
- \( u = \sin \theta \) ⇒ \( du = \cos \theta \, d\theta \)
- \( dv = e^{\sin \theta} \, d\theta \) ⇒ \( v = e^{\sin \theta} \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int e^{\sin \theta} \sin \theta \, d\theta = e^{\sin \theta} \sin \theta - \int e^{\sin \theta} \cos \theta \, d\theta \]

### Step 3: Evaluate the Second Integral
Now we need to evaluate \( \int e^{\sin \theta} \cos \theta \, d\theta \). We can use the same integration by parts method again:
- Let \( u = \cos \theta \) ⇒ \( du = -\sin \theta \, d\theta \)
- \( dv = e^{\sin \theta} \, d\theta \) ⇒ \( v = e^{\sin \theta} \)

Thus,
\[ \int e^{\sin \theta} \cos \theta \, d\theta = e^{\sin \theta} \cos \theta + \int e^{\sin \theta} \sin \theta \, d\theta \]

### Step 4: Combine the Results
Now we can substitute back into our expression:
\[ \int e^{\sin \theta} \sin \theta \, d\theta = e^{\sin \theta} \sin \theta - \left( e^{\sin \theta} \cos \theta + \int e^{\sin \theta} \sin \theta \, d\theta \right) \]
This leads to:
\[ \int e^{\sin \theta} \sin \theta \, d\theta + \int e^{\sin \theta} \sin \theta \, d\theta = e^{\sin \theta} \sin \theta - e^{\sin \theta} \cos \theta \]
\[ 2 \int e^{\sin \theta} \sin \theta \, d\theta = e^{\sin \theta} (\sin \theta - \cos \theta) \]
Thus,
\[ \int e^{\sin \theta} \sin \theta \, d\theta = \frac{1}{2} e^{\sin \theta} (\sin \theta - \cos \theta) \]

### Step 5: Evaluate the Whole Integral
Now we can evaluate the whole integral:
\[ \int e^{\sin \theta} (\sin \theta + \sec^2 \theta) \, d\theta = \frac{1}{2} e^{\sin \theta} (\sin \theta - \cos \theta) + e^{\sin \theta} \tan \theta + C \]

### Step 6: Define \( f(\theta) \)
We can define:
\[ f(\theta) = \frac{1}{2} e^{\sin \theta} (\sin \theta - \cos \theta) + e^{\sin \theta} \tan \theta \]

### Step 7: Find \( f(0) \)
To find \( f(0) \):
\[ f(0) = \frac{1}{2} e^{\sin(0)} (\sin(0) - \cos(0)) + e^{\sin(0)} \tan(0) = \frac{1}{2} (0 - 1) + 0 = -\frac{1}{2} \]
Since we need \( f(0) = 0 \), we adjust \( C \) accordingly.

### Step 8: Find \( f(\frac{\pi}{4}) \)
Now we find \( f(\frac{\pi}{4}) \):
\[ f\left(\frac{\pi}{4}\right) = \frac{1}{2} e^{\sin(\frac{\pi}{4})} \left(\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right)\right) + e^{\sin(\frac{\pi}{4})} \tan\left(\frac{\pi}{4}\right) \]
Calculating:
- \( \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
- \( \tan\left(\frac{\pi}{4}\right) = 1 \)
- \( e^{\sin\left(\frac{\pi}{4}\right)} = e^{\frac{1}{\sqrt{2}}} \)

Thus,
\[ f\left(\frac{\pi}{4}\right) = \frac{1}{2} e^{\frac{1}{\sqrt{2}}} \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) + e^{\frac{1}{\sqrt{2}}} \cdot 1 = e^{\frac{1}{\sqrt{2}}} \]

### Final Answer
The value of \( f\left(\frac{\pi}{4}\right) \) is:
\[ \boxed{e^{\frac{1}{\sqrt{2}}}} \]