Evaluate: (i) `int{1+tanxtan(x+theta)}\ dx` (ii) `int(sin2x)/(sin(x-pi/6)sin(x+pi/6))\ dx`

Explanation

Let's solve the given integrals step by step.

### Part (i): Evaluate \(\int (1 + \tan x \tan (x + \theta)) \, dx\)

1. **Rewrite the integrand**: We can use the identity for \(\tan(a + b)\):
\[ \tan(x + \theta) = \frac{\tan x + \tan \theta}{1 - \tan x \tan \theta} \]
Thus, we can express \(\tan x \tan(x + \theta)\) as:
\[ \tan x \tan(x + \theta) = \tan x \cdot \frac{\tan x + \tan \theta}{1 - \tan x \tan \theta} \]

2. **Simplify the expression**: We can rewrite \(1 + \tan x \tan(x + \theta)\) using the identity:
\[ 1 + \tan x \tan(x + \theta) = \frac{1 - \tan x \tan \theta + \tan x \tan x + \tan x \tan \theta}{1 - \tan x \tan \theta} \]
This simplifies to:
\[ \frac{1 + \tan^2 x}{1 - \tan x \tan \theta} = \frac{\sec^2 x}{1 - \tan x \tan \theta} \]

3. **Integrate**: Now we can integrate:
\[ \int \sec^2 x \, dx = \tan x + C \]
Therefore, we have:
\[ \int (1 + \tan x \tan(x + \theta)) \, dx = \tan x + C \]

### Final Answer for Part (i):
\[ \int (1 + \tan x \tan(x + \theta)) \, dx = \tan x + C \]

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### Part (ii): Evaluate \(\int \frac{\sin 2x}{\sin(x - \frac{\pi}{6}) \sin(x + \frac{\pi}{6})} \, dx\)

1. **Use the product-to-sum identities**: We know that:
\[ \sin(a - b) \sin(a + b) = \frac{1}{2} [\cos(2b) - \cos(2a)] \]
So, we can write:
\[ \sin(x - \frac{\pi}{6}) \sin(x + \frac{\pi}{6}) = \frac{1}{2} [\cos(\frac{\pi}{3}) - \cos(2x)] = \frac{1}{2} \left(\frac{1}{2} - \cos(2x)\right) \]

2. **Rewrite the integral**: This gives us:
\[ \int \frac{\sin 2x}{\frac{1}{2} \left(\frac{1}{2} - \cos(2x)\right)} \, dx = 2 \int \frac{\sin 2x}{\frac{1}{2} - \cos(2x)} \, dx \]

3. **Substitution**: Let \(t = \cos(2x)\), then \(dt = -2\sin(2x) \, dx\) or \(-\frac{1}{2} dt = \sin(2x) \, dx\). Thus, we can rewrite the integral as:
\[ -\int \frac{1}{\frac{1}{2} - t} \, dt \]

4. **Integrate**: The integral becomes:
\[ -2 \ln | \frac{1}{2} - t | + C = -2 \ln | \frac{1}{2} - \cos(2x) | + C \]

### Final Answer for Part (ii):
\[ \int \frac{\sin 2x}{\sin(x - \frac{\pi}{6}) \sin(x + \frac{\pi}{6})} \, dx = -2 \ln | \frac{1}{2} - \cos(2x) | + C \]

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