The value of `int_(0)^(pi//2) x^(10) sin x" dx"`, is then the value of `mu_(10)+90mu_(8)`, is

Options

A. `9((pi)/(2))^(8)`
B. `((pi)/(2))^(9)`
C. `10((pi)/(2))^(9)`
D. `9((pi)/(2))^(9)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} x^{10} \sin x \, dx \) and find the value of \( \mu_{10} + 90\mu_{8} \), where \( \mu_n = \int_{0}^{\frac{\pi}{2}} x^n \sin x \, dx \), we will use integration by parts.

### Step 1: Set Up Integration by Parts
We will apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = x^{10} \) (First function)
- \( dv = \sin x \, dx \) (Second function)

### Step 2: Differentiate and Integrate
Now, we differentiate \( u \) and integrate \( dv \):
- \( du = 10x^{9} \, dx \)
- \( v = -\cos x \)

### Step 3: Apply Integration by Parts
Using the integration by parts formula:
\[ I = \left[ -x^{10} \cos x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} -\cos x \cdot 10x^{9} \, dx \]
This simplifies to:
\[ I = \left[ -x^{10} \cos x \right]_{0}^{\frac{\pi}{2}} + 10 \int_{0}^{\frac{\pi}{2}} x^{9} \cos x \, dx \]

### Step 4: Evaluate the Boundary Terms
Now we evaluate the boundary terms:
\[ \left[ -x^{10} \cos x \right]_{0}^{\frac{\pi}{2}} = -\left( \left(\frac{\pi}{2}\right)^{10} \cdot 0 - 0 \cdot 1 \right) = 0 \]
Thus, we have:
\[ I = 0 + 10 \int_{0}^{\frac{\pi}{2}} x^{9} \cos x \, dx \]
Let \( J = \int_{0}^{\frac{\pi}{2}} x^{9} \cos x \, dx \).

### Step 5: Apply Integration by Parts Again
Now we need to evaluate \( J \) using integration by parts again. Let:
- \( u = x^{9} \)
- \( dv = \cos x \, dx \)

Then:
- \( du = 9x^{8} \, dx \)
- \( v = \sin x \)

Applying integration by parts:
\[ J = \left[ x^{9} \sin x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \sin x \cdot 9x^{8} \, dx \]
Evaluating the boundary terms:
\[ \left[ x^{9} \sin x \right]_{0}^{\frac{\pi}{2}} = \left( \left(\frac{\pi}{2}\right)^{9} \cdot 1 - 0 \cdot 0 \right) = \left(\frac{\pi}{2}\right)^{9} \]
Thus:
\[ J = \left(\frac{\pi}{2}\right)^{9} - 9 \int_{0}^{\frac{\pi}{2}} x^{8} \sin x \, dx \]
Let \( K = \int_{0}^{\frac{\pi}{2}} x^{8} \sin x \, dx \), then:
\[ J = \left(\frac{\pi}{2}\right)^{9} - 9K \]

### Step 6: Substitute Back
Now substitute \( J \) back into the equation for \( I \):
\[ I = 10J = 10\left(\left(\frac{\pi}{2}\right)^{9} - 9K\right) = 10\left(\frac{\pi}{2}\right)^{9} - 90K \]

### Step 7: Relate \( I \) and \( K \)
From the definition of \( \mu \):
\[ \mu_{10} + 90\mu_{8} = I \]
Thus:
\[ \mu_{10} + 90\mu_{8} = 10\left(\frac{\pi}{2}\right)^{9} \]

### Final Result
Therefore, the value of \( \mu_{10} + 90\mu_{8} \) is:
\[ \mu_{10} + 90\mu_{8} = 10\left(\frac{\pi}{2}\right)^{9} \]