Prove the equality
`int_(0)^(pi) f (sin x) dx = 2 int_(0)^(pi//2) f (sin x) dx`

Answer

To prove the equality

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

we will use a substitution and properties of definite integrals.

### Step 1: Consider the left-hand side integral

We start with the left-hand side:

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

### Step 2: Use substitution

Now, we will use the substitution \( x = \pi - t \). Therefore, when \( x = 0 \), \( t = \pi \), and when \( x = \pi \), \( t = 0 \). The differential \( dx \) becomes \( -dt \). Thus, we can rewrite the integral as follows:

\[ I = \int_{\pi}^{0} f(\sin(\pi - t)) (-dt) = \int_{0}^{\pi} f(\sin(\pi - t)) \, dt. \]

### Step 3: Simplify the sine function

Using the property of sine, we know that:

\[ \sin(\pi - t) = \sin t. \]

So, we can substitute this back into our integral:

\[ I = \int_{0}^{\pi} f(\sin t) \, dt. \]

### Step 4: Split the integral

Now we can split the integral from \(0\) to \(\pi\) into two parts:

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

### Step 5: Change of variable in the second integral

For the second integral, we can use the substitution \( x = \pi - u \). Thus, when \( x = \frac{\pi}{2} \), \( u = \frac{\pi}{2} \) and when \( x = \pi \), \( u = 0 \). The differential \( dx \) becomes \( -du \). Therefore, we have:

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

### Step 6: Combine the integrals

Now we can combine both parts:

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

### Conclusion

Thus, we have shown that:

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

which proves the equality.

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