The value of the integral `I=inte^(x)(sinx+cosx)dx` is equal to `e^(x).f(x)+C, C` being the constant of integration. Then the maximum value of `y=f(x^(2)), AA x in R` is equal to

Options

A. 0
B. -1
C. 1
D. `(1)/(2)`

Answer

To solve the integral \( I = \int e^x (\sin x + \cos x) \, dx \) and find the maximum value of \( y = f(x^2) \), where \( I = e^x f(x) + C \), we can follow these steps:

### Step 1: Identify the integral form
The integral we need to solve is:
\[ I = \int e^x (\sin x + \cos x) \, dx \]

### Step 2: Use integration by parts or recognize the pattern
We can recognize that the integral of the form \( e^x (f(x) + f'(x)) \) leads to the solution:
\[ \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C \]
In our case, we can let \( f(x) = \sin x \). Then, \( f'(x) = \cos x \).

### Step 3: Confirm the choice of \( f(x) \)
If we take \( f(x) = \sin x \), then:
\[ f'(x) = \cos x \]
Thus, \( \sin x + \cos x = f(x) + f'(x) \), which fits our integral.

### Step 4: Write the solution for the integral
Therefore, we can write:
\[ I = e^x \sin x + C \]
This means \( f(x) = \sin x \).

### Step 5: Find \( f(x^2) \)
Now we need to find \( y = f(x^2) \):
\[ y = f(x^2) = \sin(x^2) \]

### Step 6: Determine the maximum value of \( y = \sin(x^2) \)
The sine function oscillates between -1 and 1 for all real numbers. Therefore, the maximum value of \( \sin(x^2) \) is:
\[ \text{Maximum value of } y = 1 \]

### Conclusion
The maximum value of \( y = f(x^2) \) is:
\[ \boxed{1} \]