If f(x) is an integrable function in `((pi)/(6), (pi)/(3)) and I_(1)= int_(pi//6)^(pi//3) sec^(2) x f (2 sin 2x)dx and I_(2) = int_(pi//6)^(pi//3) cosec^(2) x f (2sin 2x) dx`, then:

Options

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

Answer

To solve the problem, we will analyze the integrals \( I_1 \) and \( I_2 \) and find a relationship between them.

### Step 1: Define the integrals
We have:
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^2 x \cdot f(2 \sin 2x) \, dx \]
\[ I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc^2 x \cdot f(2 \sin 2x) \, dx \]

### Step 2: Use the property of definite integrals
We will use the substitution \( x = \frac{\pi}{2} - t \). The differential \( dx \) becomes \( -dt \). The limits change as follows:
- When \( x = \frac{\pi}{6} \), \( t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \)
- When \( x = \frac{\pi}{3} \), \( t = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} \)

Thus, we can rewrite \( I_1 \):
\[ I_1 = \int_{\frac{\pi}{3}}^{\frac{\pi}{6}} \sec^2\left(\frac{\pi}{2} - t\right) f\left(2 \sin\left(2\left(\frac{\pi}{2} - t\right)\right)\right)(-dt) \]
Using the identities \( \sec\left(\frac{\pi}{2} - t\right) = \csc t \) and \( \sin\left(2\left(\frac{\pi}{2} - t\right)\right) = \sin(\pi - 2t) = \sin(2t) \), we can simplify:
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc^2 t \cdot f(2 \sin 2t) \, dt \]
This shows that:
\[ I_1 = I_2 \]

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