`int_(0)^(1) (sin theta (cos^(2) theta- cos^(2) pi//5) (cos^(2) theta - cos^(2) 2 pi //5))/(sin 5 theta)d theta`

Answer

To solve the integral

\[ I = \int_{0}^{1} \frac{\sin \theta \left( \cos^2 \theta - \cos^2 \frac{\pi}{5} \right) \left( \cos^2 \theta - \cos^2 \frac{2\pi}{5} \right)}{\sin 5\theta} \, d\theta, \]

we can use a trigonometric identity and some algebraic manipulations.

### Step 1: Use the identity for \(\sin 5\theta\)

We know that:

\[ \sin 5\theta = 16 \sin \theta \cos^4 \theta - 20 \sin \theta \cos^2 \theta + 5 \sin \theta. \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{1} \frac{\sin \theta \left( \cos^2 \theta - \cos^2 \frac{\pi}{5} \right) \left( \cos^2 \theta - \cos^2 \frac{2\pi}{5} \right)}{16 \sin \theta \cos^4 \theta - 20 \sin \theta \cos^2 \theta + 5 \sin \theta} \, d\theta. \]

### Step 2: Simplify the integral

Since \(\sin \theta\) is common in the numerator and denominator, we can cancel it out (for \(\theta \neq 0\)):

\[ I = \int_{0}^{1} \frac{\left( \cos^2 \theta - \cos^2 \frac{\pi}{5} \right) \left( \cos^2 \theta - \cos^2 \frac{2\pi}{5} \right)}{16 \cos^4 \theta - 20 \cos^2 \theta + 5} \, d\theta. \]

### Step 3: Evaluate the integral

Now, we can evaluate the integral. The integral simplifies to:

\[ I = \frac{1}{16} \int_{0}^{1} d\theta = \frac{1}{16} \cdot (1 - 0) = \frac{1}{16}. \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{\frac{1}{16}}. \]

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