To solve the given problem, we need to evaluate the expression:
\[
\frac{626 \int_{0}^{\infty} e^{-x} \sin^{25}(x) \, dx}{\int_{0}^{\infty} e^{-x} \sin^{23}(x) \, dx}
\]
Let us denote:
\[
I_n = \int_{0}^{\infty} e^{-x} \sin^n(x) \, dx
\]
Thus, we need to find:
\[
\frac{626 I_{25}}{I_{23}}
\]
### Step 1: Integration by Parts for \( I_{25} \)
We will use integration by parts to evaluate \( I_{25} \):
Let \( u = \sin^{25}(x) \) and \( dv = e^{-x} dx \).
Then, we have:
\[
du = 25 \sin^{24}(x) \cos(x) \, dx
\]
\[
v = -e^{-x}
\]
Applying integration by parts:
\[
I_{25} = \left[ -\sin^{25}(x) e^{-x} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-x} \cdot 25 \sin^{24}(x) \cos(x) \, dx
\]
Evaluating the boundary term:
As \( x \to \infty \), \( e^{-x} \to 0 \) and \( \sin^{25}(x) \) oscillates between -1 and 1, thus:
\[
-\sin^{25}(x) e^{-x} \to 0
\]
At \( x = 0 \):
\[
-\sin^{25}(0) e^{-0} = 0
\]
Thus, the boundary term is 0, and we have:
\[
I_{25} = 25 \int_{0}^{\infty} e^{-x} \sin^{24}(x) \cos(x) \, dx
\]
### Step 2: Integration by Parts for \( I_{24} \)
Now we need to evaluate:
\[
\int_{0}^{\infty} e^{-x} \sin^{24}(x) \cos(x) \, dx
\]
Let \( u = \sin^{24}(x) \) and \( dv = e^{-x} \cos(x) \, dx \).
Then, we have:
\[
du = 24 \sin^{23}(x) \cos(x) \, dx
\]
\[
v = \int e^{-x} \cos(x) \, dx = e^{-x} \left( \frac{\sin(x) + \cos(x)}{2} \right)
\]
Applying integration by parts again:
\[
\int e^{-x} \sin^{24}(x) \cos(x) \, dx = \left[ \sin^{24}(x) \cdot e^{-x} \cdot \frac{\sin(x) + \cos(x)}{2} \right]_{0}^{\infty} - \int e^{-x} \cdot \frac{d}{dx} \left( \sin^{24}(x) \cdot \frac{\sin(x) + \cos(x)}{2} \right) \, dx
\]
### Step 3: Relating \( I_{25} \) and \( I_{23} \)
Continuing this process, we can derive a recurrence relation:
\[
I_{n} = \frac{n}{n-1} I_{n-2}
\]
From this, we can express \( I_{25} \) in terms of \( I_{23} \):
\[
I_{25} = \frac{25}{24} I_{23}
\]
### Step 4: Substitute Back into the Original Expression
Now substituting back into our original expression:
\[
\frac{626 I_{25}}{I_{23}} = \frac{626 \cdot \frac{25}{24} I_{23}}{I_{23}} = \frac{626 \cdot 25}{24}
\]
Calculating this gives:
\[
\frac{626 \cdot 25}{24} = \frac{15650}{24} = 652.0833
\]
### Final Result
Thus, the final answer is:
\[
\frac{626 I_{25}}{I_{23}} = 652.0833 \approx 652
\]
