`I_(1) = int_(pi/6)^(pi/3) (dx)/(1+sqrt(tanx))` and `I_(2) = (sqrt(sinx)dx)/(sqrt(sinx) + sqrt(cosx))`
What is `I_(1) - I_(2)` equal to ?

Options

A. 0
B. `2I_(1)`
C. `pi`
D. None of the above

Explanation

To solve the problem, we need to evaluate \( I_1 \) and \( I_2 \) and then find \( I_1 - I_2 \).

### Step 1: Evaluate \( I_1 \)

Given:
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}} \]

We can rewrite \( \tan x \) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus, we have:
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\frac{\sin x}{\cos x}}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \frac{\sqrt{\sin x}}{\sqrt{\cos x}}} \]

Now, multiply the numerator and denominator by \( \sqrt{\cos x} \):
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x} \, dx}{\sqrt{\cos x} + \sqrt{\sin x}} \]

### Step 2: Use the symmetry property of definite integrals

Using the property of definite integrals:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Here \( a = \frac{\pi}{6} \) and \( b = \frac{\pi}{3} \), thus \( a + b = \frac{\pi}{2} \). We can rewrite \( I_1 \):
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos(\frac{\pi}{2} - x)} \, dx}{\sqrt{\cos(\frac{\pi}{2} - x)} + \sqrt{\sin(\frac{\pi}{2} - x)}} \]
Since \( \cos(\frac{\pi}{2} - x) = \sin x \) and \( \sin(\frac{\pi}{2} - x) = \cos x \), we have:
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} \, dx}{\sqrt{\sin x} + \sqrt{\cos x}} \]

### Step 3: Evaluate \( I_2 \)

Given:
\[ I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} \, dx}{\sqrt{\sin x} + \sqrt{\cos x}} \]

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

From the previous steps, we find that:
\[ I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x} \, dx}{\sqrt{\cos x} + \sqrt{\sin x}} \quad \text{and} \quad I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} \, dx}{\sqrt{\sin x} + \sqrt{\cos x}} \]

### Step 5: Calculate \( I_1 - I_2 \)

Since both integrals \( I_1 \) and \( I_2 \) are equal:
\[ I_1 = I_2 \]
Thus:
\[ I_1 - I_2 = 0 \]

### Final Result
\[ I_1 - I_2 = 0 \]