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 can use a property of definite integrals. This property states that if we have an integral of the form

\[ \int_{0}^{a} f(\theta) \, d\theta, \]

we can also express it as

\[ \int_{0}^{a} f(a - \theta) \, d\theta. \]

### Step 1: Apply the property to our integral

Let’s apply this property to our integral. Here, \( a = \frac{\pi}{2} \):

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

### Step 2: Simplify the expression

Now, we simplify the terms:

1. \(\sin\left(\pi - 2x\right) = \sin(2x)\)
2. \(\sin\left(\frac{\pi}{2} - x\right) = \cos x\)

Thus, we can rewrite the integral as:

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

### Step 3: Add the two expressions for \(I\)

Now we have two expressions for \(I\):

1. \(I = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx\)
2. \(I = \int_{0}^{\frac{\pi}{2}} f(\sin 2x) \cos x \, dx\)

Adding these two equations gives:

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

### Step 4: Factor out the common terms

Now, we can factor out the common terms:

\[ I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} f(\sin 2x) (\sin x + \cos x) \, dx. \]

### Step 5: Use the identity for \(\sin x + \cos x\)

Notice that \(\sin x + \cos x\) can be rewritten using the identity:

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

Thus, we can express our integral as:

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

### Step 6: Final expression for \(I\)

This leads us to the final expression for \(I\):

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

### Conclusion

Thus, the integral

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

is equal to

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