Consider a real valued continuous function `f` such that `f(x)=sinx + int_(-pi//2)^(pi//2) (sinx+t(f(t))dt`. If `M` and `m` are maximum and minimum values of function `f`, then the value of `M//m` is____________.

Explanation

To solve the problem, we start with the function defined as:

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

### Step 1: Simplify the Integral
We can break down the integral into two parts:

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

The first integral simplifies because \(\sin x\) is constant with respect to \(t\):

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin x \, dt = \sin x \cdot \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = \sin x \cdot \pi \]

Thus, we have:

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

### Step 2: Combine Terms
Now we can combine the terms involving \(\sin x\):

\[ f(x) = (1 + \pi) \sin x + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t f(t) \, dt \]

Let \(A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t f(t) \, dt\). Then we can write:

\[ f(x) = (1 + \pi) \sin x + A \]

### Step 3: Determine the Value of A
To find \(A\), we can substitute \(f(t)\) back into the integral:

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

This simplifies to:

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

The second integral, \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \, dt\), evaluates to 0 because it is symmetric around 0. Therefore, we have:

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

### Step 4: Calculate the Integral
Using integration by parts for \(\int t \sin t \, dt\):

Let \(u = t\) and \(dv = \sin t \, dt\). Then \(du = dt\) and \(v = -\cos t\).

Using integration by parts:

\[ \int t \sin t \, dt = -t \cos t + \int \cos t \, dt = -t \cos t + \sin t \]

Evaluating from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\):

\[ \left[-t \cos t + \sin t\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \left[-\frac{\pi}{2} \cdot 0 + 1\right] - \left[\frac{\pi}{2} \cdot 0 - 1\right] = 1 - (-1) = 2 \]

Thus, \(A = (1 + \pi) \cdot 2 = 2(1 + \pi)\).

### Step 5: Substitute A Back into f(x)
Now substituting \(A\) back into \(f(x)\):

\[ f(x) = (1 + \pi) \sin x + 2(1 + \pi) \]

### Step 6: Find Maximum and Minimum Values
The maximum value of \(\sin x\) is 1, and the minimum value is -1. Therefore:

- Maximum \(M\) of \(f(x)\):

\[ M = (1 + \pi) \cdot 1 + 2(1 + \pi) = 3 + 3\pi \]

- Minimum \(m\) of \(f(x)\):

\[ m = (1 + \pi) \cdot (-1) + 2(1 + \pi) = -1 - \pi + 2 + 2\pi = 1 + \pi \]

### Step 7: Calculate \(\frac{M}{m}\)
Now we calculate \(\frac{M}{m}\):

\[ \frac{M}{m} = \frac{3 + 3\pi}{1 + \pi} \]

### Final Result
Thus, the value of \(\frac{M}{m}\) is:

\[ \frac{M}{m} = 3 \]