To prove the equality of the three integrals given in the question, we will denote the integrals as follows:
Let:
- \( I_1 = \int_{0}^{\pi} x f(\sin x) \, dx \)
- \( I_2 = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \)
- \( I_3 = \pi \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \)
We will show that \( I_1 = I_2 = I_3 \).
### Step 1: Evaluate \( I_1 \)
Using the property of definite integrals, we have:
\[
\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx
\]
For our case, we set \( a = \pi \). Therefore:
\[
I_1 = \int_{0}^{\pi} x f(\sin x) \, dx = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x)) \, dx
\]
Since \( \sin(\pi - x) = \sin x \), we can rewrite this as:
\[
I_1 = \int_{0}^{\pi} (\pi - x) f(\sin x) \, dx
\]
### Step 2: Simplify \( I_1 \)
Now, we can split \( I_1 \):
\[
I_1 = \int_{0}^{\pi} \pi f(\sin x) \, dx - \int_{0}^{\pi} x f(\sin x) \, dx
\]
This gives us:
\[
I_1 = \pi \int_{0}^{\pi} f(\sin x) \, dx - I_1
\]
Adding \( I_1 \) to both sides:
\[
2I_1 = \pi \int_{0}^{\pi} f(\sin x) \, dx
\]
Dividing by 2:
\[
I_1 = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx
\]
Thus, we have shown that:
\[
I_1 = I_2
\]
### Step 3: Show \( I_2 = I_3 \)
Now, we will show that \( I_2 = I_3 \). We use the property of definite integrals again:
\[
\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(a - x) \, dx
\]
Setting \( a = \frac{\pi}{2} \):
\[
I_2 = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx
\]
We can change the limits of integration for \( I_3 \):
\[
I_3 = \pi \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx
\]
Using the substitution \( u = \sin x \), we find:
\[
du = \cos x \, dx \quad \text{and} \quad dx = \frac{du}{\sqrt{1-u^2}}
\]
The limits change from \( 0 \) to \( 1 \) as \( x \) goes from \( 0 \) to \( \frac{\pi}{2} \).
Thus:
\[
I_3 = \pi \int_{0}^{1} f(u) \, du
\]
Now, since \( \int_{0}^{\pi} f(\sin x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \), we can conclude:
\[
I_2 = I_3
\]
### Conclusion
We have shown that:
\[
I_1 = I_2 = I_3
\]
Thus, the original statement is proved.
