`int_(0)^(pi//2) f(sin 2x) sin x dx= int_(0)^(pi//2) f(sin 2x) cos x dx= sqrt""2 int_(0)^(pi//4) f(cos 2x) cos x dx`

Answer

To solve the equation

\[ \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \cos x \, dx = \sqrt{2} \int_{0}^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx, \]

we will denote the integrals as follows:

- Let \( I_1 = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx \)
- Let \( I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \cos x \, dx \)
- Let \( I_3 = \int_{0}^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx \)

### Step 1: Show that \( I_1 = I_2 \)

Using the property of integrals:

\[ \int_{0}^{A} f(x) \, dx = \int_{0}^{A} f(A - x) \, dx, \]

we can transform \( I_1 \):

\[ I_1 = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx = \int_{0}^{\frac{\pi}{2}} f(\sin(2(\frac{\pi}{2} - x))) \sin(\frac{\pi}{2} - x) \, dx. \]

Since \( \sin(2(\frac{\pi}{2} - x)) = \sin(\pi - 2x) = \sin(2x) \) and \( \sin(\frac{\pi}{2} - x) = \cos x \), we have:

\[ I_1 = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \cos x \, dx = I_2. \]

### Step 2: Show that \( I_1 + I_2 = 2I_1 \)

We can add \( I_1 \) and \( I_2 \):

\[ I_1 + I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) (\sin x + \cos x) \, dx. \]

### Step 3: Simplify \( \sin x + \cos x \)

Using the identity:

\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right), \]

we can rewrite the integral:

\[ I_1 + I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \, dx. \]

### Step 4: Change of variables for \( I_3 \)

Now, we will relate \( I_3 \) to \( I_1 \) and \( I_2 \). We can use the substitution \( x = \frac{\pi}{4} - u \) in \( I_3 \):

\[ I_3 = \int_{0}^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx. \]

Using the property of integrals again, we can show that:

\[ I_3 = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{4}} f(\sin 2x) \, dx. \]

### Conclusion

From the above steps, we have shown that:

\[ I_1 = I_2 = \sqrt{2} I_3. \]

Thus, we conclude that:

\[ I_1 = I_2 = I_3. \]