`f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt`
The value of `int_(0)^(pi//2) f(x)dx` is

Options

A. `1`
B. `-2`
C. `-1`
D. `2`

Answer

To solve the problem, we need to find the value of the integral \( \int_{0}^{\frac{\pi}{2}} f(x) \, dx \) where

\[ f(x) = \sin x + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin x + t \cos x) f(t) \, dt. \]

### Step 1: Rewrite the function \( f(x) \)

We can express \( f(x) \) as:

\[ f(x) = \sin x + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin x + t \cos x) f(t) \, dt. \]

This can be simplified to:

\[ f(x) = \sin x + \sin x \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(t) \, dt + \cos x \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t f(t) \, dt. \]

Let \( A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(t) \, dt \) and \( B = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t f(t) \, dt \). Then we can rewrite \( f(x) \) as:

\[ f(x) = \sin x (1 + A) + \cos x B. \]

### Step 2: Determine constants \( A \) and \( B \)

1. **Finding \( A \)**:

Substitute \( f(t) = \sin t (1 + A) + \cos t B \) into the integral for \( A \):

\[ A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sin t (1 + A) + \cos t B \right) dt. \]

This separates into:

\[ A = (1 + A) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin t \, dt + B \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos t \, dt. \]

The integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin t \, dt = 0 \) and \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos t \, dt = 2 \).

Thus,

\[ A = 0 + 2B \implies A = 2B. \]

2. **Finding \( B \)**:

Substitute \( f(t) \) into the integral for \( B \):

\[ B = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \left( \sin t (1 + A) + \cos t B \right) dt. \]

This separates into:

\[ B = (1 + A) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \sin t \, dt + B \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \cos t \, dt. \]

The integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \sin t \, dt = 0 \) (since it is an odd function) and \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \cos t \, dt \) can be computed using integration by parts, yielding \( 0 \).

Thus,

\[ B = 0 + 0 \implies B = 0. \]

### Step 3: Substitute back to find \( A \)

Since \( B = 0 \), we have:

\[ A = 2B = 0. \]

### Step 4: Final form of \( f(x) \)

Now substituting \( A \) and \( B \) back into the expression for \( f(x) \):

\[ f(x) = \sin x (1 + 0) + \cos x (0) = \sin x. \]

### Step 5: Calculate the integral

Now we can compute:

\[ \int_{0}^{\frac{\pi}{2}} f(x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin x \, dx. \]

The integral of \( \sin x \) is:

\[ -\cos x \bigg|_{0}^{\frac{\pi}{2}} = -\left(0 - (-1)\right) = 1. \]

### Conclusion

Thus, the value of \( \int_{0}^{\frac{\pi}{2}} f(x) \, dx \) is:

\[ \boxed{1}. \]