`int_(pi//6)^(pi//3)cos^(2)x dx=`

Options

A. `(pi)/(6)`
B. `(pi)/(12)`
C. `(sqrt(3)-1)/(2)`
D. `(sqrt(3)+1)/(2)`

Answer

To solve the definite integral \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 x \, dx\), we can use the property of definite integrals which states that:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

### Step 1: Set up the integral
Let \( I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 x \, dx \).

### Step 2: Apply the property of definite integrals
Using the property mentioned above, we can rewrite the integral:

\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2\left(\frac{\pi}{2} - x\right) \, dx \]

### Step 3: Simplify using trigonometric identities
We know that:

\[ \cos\left(\frac{\pi}{2} - x\right) = \sin x \]

Thus,

\[ \cos^2\left(\frac{\pi}{2} - x\right) = \sin^2 x \]

So we can rewrite the integral as:

\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \, dx \]

### Step 4: Combine the two integrals
Now we have two expressions for \( I \):

\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 x \, dx \]
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \, dx \]

Adding these two equations gives:

\[ 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left(\cos^2 x + \sin^2 x\right) \, dx \]

### Step 5: Use the Pythagorean identity
Using the identity \(\cos^2 x + \sin^2 x = 1\):

\[ 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx \]

### Step 6: Evaluate the integral
Now we can evaluate the integral:

\[ 2I = \left[x\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} \]

Calculating this gives:

\[ 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6} \]

### Step 7: Solve for \( I \)
Now, divide both sides by 2:

\[ I = \frac{\pi}{6} \cdot \frac{1}{2} = \frac{\pi}{12} \]

### Final Answer
Thus, the value of the integral is:

\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 x \, dx = \frac{\pi}{12} \]

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