The integral `I=int_((pi)/(4))^((pi)/(2))sin^(6)xdx` is satisfies

Options

A. `I gt (pi)/(2)`
B. `I gt (pi)/(4)`
C. `I in ((pi)/(4), (pi)/(2))`
D. `I in ((pi)/(32), (pi)/(4))`

Answer

To solve the integral \( I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^6 x \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
We start with the integral:
\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^6 x \, dx \]

2. **Use the Power Reduction Formula**:
We can express \(\sin^6 x\) using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\):
\[ \sin^6 x = (\sin^2 x)^3 = \left(\frac{1 - \cos 2x}{2}\right)^3 \]
Expanding this gives:
\[ \sin^6 x = \frac{(1 - \cos 2x)^3}{8} \]

3. **Expand the Cubic**:
Now we expand \((1 - \cos 2x)^3\):
\[ (1 - \cos 2x)^3 = 1 - 3\cos 2x + 3\cos^2 2x - \cos^3 2x \]
Using \(\cos^2 2x = \frac{1 + \cos 4x}{2}\) and \(\cos^3 2x = \frac{3\cos 2x + \cos 6x}{4}\), we can substitute:
\[ \cos^2 2x = \frac{1 + \cos 4x}{2} \]
Thus:
\[ (1 - \cos 2x)^3 = 1 - 3\cos 2x + \frac{3(1 + \cos 4x)}{2} - \frac{3\cos 2x + \cos 6x}{4} \]

4. **Combine Terms**:
Combine all terms together:
\[ = 1 - 3\cos 2x + \frac{3}{2} + \frac{3}{2}\cos 4x - \frac{3\cos 2x}{4} - \frac{\cos 6x}{4} \]
Collecting like terms gives:
\[ = \frac{5}{2} - \left(3 + \frac{3}{4}\right)\cos 2x + \frac{3}{2}\cos 4x - \frac{1}{4}\cos 6x \]

5. **Integrate**:
Now we substitute this back into the integral:
\[ I = \frac{1}{8} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left(\frac{5}{2} - \frac{15}{4}\cos 2x + \frac{3}{2}\cos 4x - \frac{1}{4}\cos 6x\right) dx \]
Integrating term by term:
- The integral of a constant:
\[ \int dx = x \]
- The integral of \(\cos kx\):
\[ \int \cos kx \, dx = \frac{\sin kx}{k} \]

6. **Evaluate the Integral**:
Evaluating each term from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\):
\[ I = \frac{1}{8} \left[ \frac{5}{2}x - \frac{15}{4} \cdot \frac{\sin 2x}{2} + \frac{3}{2} \cdot \frac{\sin 4x}{4} - \frac{1}{4} \cdot \frac{\sin 6x}{6} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]

7. **Calculate the Limits**:
Substitute the limits:
- For \(x = \frac{\pi}{2}\):
\[ \sin 2\left(\frac{\pi}{2}\right) = \sin \pi = 0, \quad \sin 4\left(\frac{\pi}{2}\right) = \sin 2\pi = 0, \quad \sin 6\left(\frac{\pi}{2}\right) = \sin 3\pi = 0 \]
- For \(x = \frac{\pi}{4}\):
\[ \sin 2\left(\frac{\pi}{4}\right) = \sin \frac{\pi}{2} = 1, \quad \sin 4\left(\frac{\pi}{4}\right) = \sin \pi = 0, \quad \sin 6\left(\frac{\pi}{4}\right) = \sin \frac{3\pi}{2} = -1 \]

8. **Final Calculation**:
\[ I = \frac{1}{8} \left[ 0 - \left(\frac{5}{2} \cdot \frac{\pi}{4} - \frac{15}{4} \cdot \frac{1}{2} + 0 + \frac{1}{24}\right) \right] \]
Simplifying gives:
\[ I = \frac{1}{8} \left[ -\frac{5\pi}{8} + \frac{15}{8} + \frac{1}{24} \right] \]
After simplifying, we find:
\[ I = \frac{11}{48} \]

### Final Answer:
\[ I = \frac{11}{48} \]