`int_(0)^(pi//2)sin^(4)xcos^(5)x dx=`

Options

A. `(8)/(315)`
B. `(315)/(8)`
C. `(256)/(5)`
D. `(225)/(8)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^5 x \, dx \), we will follow a systematic approach.

### Step-by-step Solution:

1. **Rewrite the integral**:
We can express \( \cos^5 x \) as \( \cos^4 x \cdot \cos x \). Thus, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^4 x \cos x \, dx \]

2. **Use the identity for \( \cos^4 x \)**:
We know that \( \cos^4 x = (\cos^2 x)^2 = (1 - \sin^2 x)^2 \). Therefore, we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^4 x (1 - \sin^2 x)^2 \cos x \, dx \]

3. **Expand the integrand**:
Expanding \( (1 - \sin^2 x)^2 \) gives:
\[ (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x \]
Thus, the integral becomes:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^4 x (1 - 2\sin^2 x + \sin^4 x) \cos x \, dx \]
This can be split into three separate integrals:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos x \, dx - 2 \int_{0}^{\frac{\pi}{2}} \sin^6 x \cos x \, dx + \int_{0}^{\frac{\pi}{2}} \sin^8 x \cos x \, dx \]

4. **Substitution**:
Let \( t = \sin x \). Then, \( dt = \cos x \, dx \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{2} \) which corresponds to \( t = 0 \) to \( t = 1 \). The integral becomes:
\[ I = \int_{0}^{1} t^4 \, dt - 2 \int_{0}^{1} t^6 \, dt + \int_{0}^{1} t^8 \, dt \]

5. **Evaluate each integral**:
Using the formula \( \int t^n \, dt = \frac{t^{n+1}}{n+1} \):
- \( \int_{0}^{1} t^4 \, dt = \frac{1^5}{5} = \frac{1}{5} \)
- \( \int_{0}^{1} t^6 \, dt = \frac{1^7}{7} = \frac{1}{7} \)
- \( \int_{0}^{1} t^8 \, dt = \frac{1^9}{9} = \frac{1}{9} \)

6. **Combine the results**:
Substituting back into the expression for \( I \):
\[ I = \frac{1}{5} - 2 \cdot \frac{1}{7} + \frac{1}{9} \]
Finding a common denominator (which is 315):
\[ I = \frac{63}{315} - \frac{90}{315} + \frac{35}{315} = \frac{63 - 90 + 35}{315} = \frac{8}{315} \]

### Final Answer:
Thus, the value of the integral is:
\[ I = \frac{8}{315} \]