If f(x) is an integrable function on `[(pi)/(6),(pi)/(3)]` and
`I_(1)=int_(pi//6)^(pi//3) sec^(2)thetaf(2 sin 2theta)d theta" and "I_(2)=int_(pi//6)^(pi//3) "cosec"^(2)theta f(2 sin2theta)d theta`, then

Options

A. `I_(1)=2I_(2)`
B. `I_(1)=3I_(2)`
C. `2I_(1)=I_(2)`
D. none of these

Explanation

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and establish a relationship between them.

Given:
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^2 \theta \, f(2 \sin 2\theta) \, d\theta \]
\[ I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc^2 \theta \, f(2 \sin 2\theta) \, d\theta \]

### Step 1: Use the property of definite integrals
We can use the property of definite integrals which states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Here, we will let \( a = \frac{\pi}{6} \) and \( b = \frac{\pi}{3} \).

### Step 2: Substitute \( \theta \) in \( I_1 \)
Using the property, we can substitute \( \theta \) with \( \frac{\pi}{2} - \theta \) in \( I_1 \):
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^2\left(\frac{\pi}{2} - \theta\right) f\left(2 \sin\left(2\left(\frac{\pi}{2} - \theta\right)\right)\right) d\theta \]

### Step 3: Simplify the integrand
Using trigonometric identities:
- \( \sec^2\left(\frac{\pi}{2} - \theta\right) = \csc^2 \theta \)
- \( \sin(2(\frac{\pi}{2} - \theta)) = \sin(\pi - 2\theta) = \sin(2\theta) \)

Thus, we can rewrite \( I_1 \):
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc^2 \theta \, f(2 \sin 2\theta) \, d\theta \]

### Step 4: Compare \( I_1 \) and \( I_2 \)
Now we see that:
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc^2 \theta \, f(2 \sin 2\theta) \, d\theta = I_2 \]

### Conclusion
Thus, we have established that:
\[ I_1 = I_2 \]

### Final Result
The relationship between \( I_1 \) and \( I_2 \) is:
\[ I_1 = I_2 \]