To evaluate the integral \( \int_0^{\frac{\pi}{6}} \cos x \cos 2x \, dx \), we can use the product-to-sum identities. Here’s a step-by-step solution:
### Step 1: Use the Product-to-Sum Identity
We know the identity:
\[
\cos A \cos B = \frac{1}{2} (\cos(A + B) + \cos(A - B))
\]
For our integral, let \( A = x \) and \( B = 2x \). Thus, we have:
\[
\cos x \cos 2x = \frac{1}{2} (\cos(x + 2x) + \cos(x - 2x)) = \frac{1}{2} (\cos 3x + \cos x)
\]
### Step 2: Rewrite the Integral
Substituting this identity into the integral, we get:
\[
\int_0^{\frac{\pi}{6}} \cos x \cos 2x \, dx = \int_0^{\frac{\pi}{6}} \frac{1}{2} (\cos 3x + \cos x) \, dx
\]
This simplifies to:
\[
= \frac{1}{2} \int_0^{\frac{\pi}{6}} (\cos 3x + \cos x) \, dx
\]
### Step 3: Split the Integral
Now we can split the integral:
\[
= \frac{1}{2} \left( \int_0^{\frac{\pi}{6}} \cos 3x \, dx + \int_0^{\frac{\pi}{6}} \cos x \, dx \right)
\]
### Step 4: Evaluate Each Integral
1. **For \( \int_0^{\frac{\pi}{6}} \cos 3x \, dx \)**:
\[
\int \cos 3x \, dx = \frac{1}{3} \sin 3x
\]
Evaluating from \( 0 \) to \( \frac{\pi}{6} \):
\[
= \left[ \frac{1}{3} \sin 3x \right]_0^{\frac{\pi}{6}} = \frac{1}{3} \sin\left(3 \cdot \frac{\pi}{6}\right) - \frac{1}{3} \sin(0) = \frac{1}{3} \sin\left(\frac{\pi}{2}\right) - 0 = \frac{1}{3}(1) = \frac{1}{3}
\]
2. **For \( \int_0^{\frac{\pi}{6}} \cos x \, dx \)**:
\[
\int \cos x \, dx = \sin x
\]
Evaluating from \( 0 \) to \( \frac{\pi}{6} \):
\[
= \left[ \sin x \right]_0^{\frac{\pi}{6}} = \sin\left(\frac{\pi}{6}\right) - \sin(0) = \frac{1}{2} - 0 = \frac{1}{2}
\]
### Step 5: Combine the Results
Now substituting back into our expression:
\[
= \frac{1}{2} \left( \frac{1}{3} + \frac{1}{2} \right)
\]
Finding a common denominator (which is 6):
\[
= \frac{1}{2} \left( \frac{2}{6} + \frac{3}{6} \right) = \frac{1}{2} \left( \frac{5}{6} \right) = \frac{5}{12}
\]
### Final Answer
Thus, the value of the integral is:
\[
\int_0^{\frac{\pi}{6}} \cos x \cos 2x \, dx = \frac{5}{12}
\]

Evaluate : `int_0^(pi/6)cosxcos2xdx`

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