`int_(0)^((pi)/2)cos^(6)x dx=`

Options

A. `(5pi)/16`
B. `(5pi)/32`
C. `5/16`
D. `5/32`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \cos^6 x \, dx \), we can use the reduction formula for integrals of the form \( \int_{0}^{\frac{\pi}{2}} \cos^n x \, dx \).

### Step-by-Step Solution:

1. **Identify the Integral**:
We need to evaluate:
\[ I = \int_{0}^{\frac{\pi}{2}} \cos^6 x \, dx \]

2. **Use the Reduction Formula**:
The reduction formula for the integral of \( \cos^n x \) is given by:
\[ \int_{0}^{\frac{\pi}{2}} \cos^n x \, dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \cos^{n-2} x \, dx \]
For our case, \( n = 6 \):
\[ I = \frac{6-1}{6} \int_{0}^{\frac{\pi}{2}} \cos^4 x \, dx = \frac{5}{6} I_4 \]

3. **Calculate \( I_4 \)**:
Now we need to find \( I_4 = \int_{0}^{\frac{\pi}{2}} \cos^4 x \, dx \):
\[ I_4 = \frac{4-1}{4} \int_{0}^{\frac{\pi}{2}} \cos^2 x \, dx = \frac{3}{4} I_2 \]

4. **Calculate \( I_2 \)**:
Next, we find \( I_2 = \int_{0}^{\frac{\pi}{2}} \cos^2 x \, dx \):
\[ I_2 = \frac{2-1}{2} \int_{0}^{\frac{\pi}{2}} \cos^0 x \, dx = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]

5. **Substitute Back**:
Now substitute \( I_2 \) back into the equation for \( I_4 \):
\[ I_4 = \frac{3}{4} \cdot \frac{\pi}{4} = \frac{3\pi}{16} \]

6. **Final Calculation for \( I \)**:
Substitute \( I_4 \) back into the equation for \( I \):
\[ I = \frac{5}{6} \cdot \frac{3\pi}{16} = \frac{15\pi}{96} = \frac{5\pi}{32} \]

### Final Result:
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \cos^6 x \, dx = \frac{5\pi}{32} \]