The integral `int_(0)^(pi//2) f(sin 2 x)sin x dx` is equal to

Options

A. `overset(pi//2)underset(0)int f(cos 2 x)sin x dx =sqrt(2)overset(pi//2)underset(0)int f(cos 2x)sin x dx`
B. `overset(pi//2)underset(0)int f(sin 2 x)cos x dx =sqrt(2)overset(pi//4)underset(0)int f(cos 2x)cos x dx`
C. `overset(pi//2)underset(0)int f(cos 2 x)cos x dx =sqrt(2)overset(pi//2)underset(0)int f(cos 2x)cos x dx`
D. `overset(pi//2)underset(0)int f(sin 2 x)cos x dx =sqrt(2)overset(pi//2)underset(0)int f(cos 2x)cos x dx`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx \), we will use the property of definite integrals.

### Step-by-Step Solution:

1. **Apply the property of definite integrals**:
We know that for any function \( f \), the following property holds:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
Here, we will set \( a = \frac{\pi}{2} \). Thus, we can rewrite our integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx = \int_{0}^{\frac{\pi}{2}} f(\sin 2(\frac{\pi}{2} - x)) \sin(\frac{\pi}{2} - x) \, dx \]

2. **Simplify the expression**:
Using the identities \( \sin(\frac{\pi}{2} - x) = \cos x \) and \( \sin 2(\frac{\pi}{2} - x) = \sin(2x) \), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \cos x \, dx \]

3. **Combine the two expressions**:
Now we have two expressions for \( I \):
\[ I = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx \quad \text{and} \quad I = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \cos x \, dx \]
Adding these two equations gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) (\sin x + \cos x) \, dx \]

4. **Factor out constants**:
We can factor out \( \sqrt{2} \) from \( \sin x + \cos x \):
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
Thus, we have:
\[ 2I = \sqrt{2} \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin\left(x + \frac{\pi}{4}\right) \, dx \]

5. **Final expression for \( I \)**:
Dividing both sides by 2 gives:
\[ I = \frac{\sqrt{2}}{2} \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin\left(x + \frac{\pi}{4}\right) \, dx \]

### Conclusion:
The integral \( I \) can be expressed in terms of the function \( f \) and a transformed sine function. The exact value of \( I \) will depend on the specific form of \( f \).