`int_(0)^(pi//8) cos^(3)4 theta d theta=`

Options

A. `2/3`
B. `1/4`
C. `1/3`
D. `1/6`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{8}} \cos^3(4\theta) \, d\theta \), we can follow these steps:

### Step 1: Use the identity for cosine cubed
We can use the identity for \( \cos^3 x \):
\[ \cos^3 x = \frac{1}{4} \cos(3x) + \frac{3}{4} \cos x \]
Applying this identity, we rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{8}} \cos^3(4\theta) \, d\theta = \int_{0}^{\frac{\pi}{8}} \left( \frac{1}{4} \cos(12\theta) + \frac{3}{4} \cos(4\theta) \right) d\theta \]

### Step 2: Split the integral
Now we can split the integral into two parts:
\[ I = \frac{1}{4} \int_{0}^{\frac{\pi}{8}} \cos(12\theta) \, d\theta + \frac{3}{4} \int_{0}^{\frac{\pi}{8}} \cos(4\theta) \, d\theta \]

### Step 3: Evaluate the first integral
The integral of \( \cos(k\theta) \) is given by:
\[ \int \cos(k\theta) \, d\theta = \frac{1}{k} \sin(k\theta) \]
Thus, for the first integral:
\[ \int_{0}^{\frac{\pi}{8}} \cos(12\theta) \, d\theta = \left[ \frac{1}{12} \sin(12\theta) \right]_{0}^{\frac{\pi}{8}} = \frac{1}{12} \sin\left(12 \cdot \frac{\pi}{8}\right) - \frac{1}{12} \sin(0) = \frac{1}{12} \sin\left(\frac{3\pi}{2}\right) = \frac{1}{12} \cdot (-1) = -\frac{1}{12} \]

### Step 4: Evaluate the second integral
Now for the second integral:
\[ \int_{0}^{\frac{\pi}{8}} \cos(4\theta) \, d\theta = \left[ \frac{1}{4} \sin(4\theta) \right]_{0}^{\frac{\pi}{8}} = \frac{1}{4} \sin\left(4 \cdot \frac{\pi}{8}\right) - \frac{1}{4} \sin(0) = \frac{1}{4} \sin\left(\frac{\pi}{2}\right) = \frac{1}{4} \cdot 1 = \frac{1}{4} \]

### Step 5: Substitute back into the expression for \( I \)
Now substituting back into the expression for \( I \):
\[ I = \frac{1}{4} \left(-\frac{1}{12}\right) + \frac{3}{4} \left(\frac{1}{4}\right) = -\frac{1}{48} + \frac{3}{16} \]

### Step 6: Find a common denominator and simplify
To combine these fractions, we find a common denominator (48):
\[ -\frac{1}{48} + \frac{3 \cdot 3}{48} = -\frac{1}{48} + \frac{9}{48} = \frac{8}{48} = \frac{1}{6} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{1}{6} \]