If `I_(1)=int_(0)^(pi//2)f(sin2x)sin x dx` and `I_(2)=int_(0)^(pi//4)f(cos2x)cosx dx`, then `I_(1)//I_(2)` is equal to

Options

A. 1
B. 2
C. `1//sqrt(2)`
D. `sqrt(2)`

Answer

To solve the problem, we need to find the ratio \( \frac{I_1}{I_2} \) where:

\[ I_1 = \int_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx \]
\[ I_2 = \int_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx \]

### Step 1: Change of Variable for \( I_1 \)

Let's first evaluate \( I_1 \). We can use the substitution \( x = \frac{\pi}{2} - t \). This gives us:

\[ dx = -dt \]

When \( x = 0 \), \( t = \frac{\pi}{2} \) and when \( x = \frac{\pi}{2} \), \( t = 0 \). Thus, we can rewrite \( I_1 \):

\[ I_1 = \int_{\frac{\pi}{2}}^0 f(\sin(2(\frac{\pi}{2} - t))) \sin(\frac{\pi}{2} - t)(-dt) \]

This simplifies to:

\[ I_1 = \int_0^{\frac{\pi}{2}} f(\sin(\pi - 2t)) \cos t \, dt \]

Using the identity \( \sin(\pi - x) = \sin x \):

\[ I_1 = \int_0^{\frac{\pi}{2}} f(\sin(2t)) \cos t \, dt \]

### Step 2: Break Down \( I_1 \)

We can break \( I_1 \) into two parts:

\[ I_1 = \int_0^{\frac{\pi}{4}} f(\sin(2t)) \cos t \, dt + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin(2t)) \cos t \, dt \]

### Step 3: Change of Variable for \( I_2 \)

Now, let's evaluate \( I_2 \):

Using the substitution \( x = t \), we have:

\[ I_2 = \int_0^{\frac{\pi}{4}} f(\cos(2t)) \cos t \, dt \]

### Step 4: Relate \( I_1 \) and \( I_2 \)

We can relate \( I_1 \) and \( I_2 \) by noticing that both integrals involve \( f \) evaluated at different arguments. We can express \( I_1 \) in terms of \( I_2 \):

From the previous steps, we have:

\[ I_1 = \int_0^{\frac{\pi}{4}} f(\sin(2t)) \cos t \, dt + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin(2t)) \cos t \, dt \]

Using properties of sine and cosine, we can show that:

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

### Step 5: Calculate \( \frac{I_1}{I_2} \)

Now, we can find the ratio:

\[ \frac{I_1}{I_2} = \frac{\sqrt{2} I_2}{I_2} = \sqrt{2} \]

### Final Answer

Thus, the final answer is:

\[ \frac{I_1}{I_2} = \sqrt{2} \]