If `I_(n)=int_(0)^(pi//2) x^(n) sin x dx`, then `I_(4)+12I_(2)` is equal to\

Options

A. `4pi`
B. `3((pi)/(2))^(3)`
C. `((pi)/(2))^(2)`
D. `4((pi)/(2))^(3)`

Answer

To solve the problem, we need to evaluate the expression \( I_4 + 12I_2 \) where \( I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx \).

### Step 1: Calculate \( I_4 \)

We start with:
\[ I_4 = \int_0^{\frac{\pi}{2}} x^4 \sin x \, dx \]

Using integration by parts, we let:
- \( u = x^4 \) (first function)
- \( dv = \sin x \, dx \) (second function)

Then, we differentiate and integrate:
- \( du = 4x^3 \, dx \)
- \( v = -\cos x \)

Applying integration by parts:
\[ I_4 = \left[ -x^4 \cos x \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} -\cos x \cdot 4x^3 \, dx \]

### Step 2: Evaluate the boundary term

Calculating the boundary term:
\[ \left[ -x^4 \cos x \right]_0^{\frac{\pi}{2}} = -\left( -\frac{\pi^4}{16} \cdot 0 \right) - \left( -0^4 \cdot 1 \right) = 0 \]

So we have:
\[ I_4 = 0 + 4 \int_0^{\frac{\pi}{2}} x^3 \cos x \, dx \]

### Step 3: Calculate \( \int_0^{\frac{\pi}{2}} x^3 \cos x \, dx \)

Now we need to evaluate:
\[ \int_0^{\frac{\pi}{2}} x^3 \cos x \, dx \]

Again using integration by parts:
- Let \( u = x^3 \) and \( dv = \cos x \, dx \)
- Then \( du = 3x^2 \, dx \) and \( v = \sin x \)

Applying integration by parts:
\[ \int_0^{\frac{\pi}{2}} x^3 \cos x \, dx = \left[ x^3 \sin x \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} 3x^2 \sin x \, dx \]

### Step 4: Evaluate the boundary term again

Calculating the boundary term:
\[ \left[ x^3 \sin x \right]_0^{\frac{\pi}{2}} = \left( \frac{\pi^3}{8} \cdot 1 \right) - (0 \cdot 0) = \frac{\pi^3}{8} \]

So we have:
\[ \int_0^{\frac{\pi}{2}} x^3 \cos x \, dx = \frac{\pi^3}{8} - 3 \int_0^{\frac{\pi}{2}} x^2 \sin x \, dx \]

### Step 5: Calculate \( I_2 \)

Next, we need to calculate \( I_2 = \int_0^{\frac{\pi}{2}} x^2 \sin x \, dx \) using integration by parts:
- Let \( u = x^2 \) and \( dv = \sin x \, dx \)
- Then \( du = 2x \, dx \) and \( v = -\cos x \)

Applying integration by parts:
\[ I_2 = \left[ -x^2 \cos x \right]_0^{\frac{\pi}{2}} + 2 \int_0^{\frac{\pi}{2}} x \cos x \, dx \]

### Step 6: Evaluate the boundary term for \( I_2 \)

Calculating the boundary term:
\[ \left[ -x^2 \cos x \right]_0^{\frac{\pi}{2}} = 0 \]

So we have:
\[ I_2 = 2 \int_0^{\frac{\pi}{2}} x \cos x \, dx \]

### Step 7: Calculate \( \int_0^{\frac{\pi}{2}} x \cos x \, dx \)

Using integration by parts again:
- Let \( u = x \) and \( dv = \cos x \, dx \)
- Then \( du = dx \) and \( v = \sin x \)

Applying integration by parts:
\[ \int_0^{\frac{\pi}{2}} x \cos x \, dx = \left[ x \sin x \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \sin x \, dx \]

Calculating the boundary term:
\[ \left[ x \sin x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \cdot 1 - 0 = \frac{\pi}{2} \]

So we have:
\[ \int_0^{\frac{\pi}{2}} x \cos x \, dx = \frac{\pi}{2} - 1 \]

### Step 8: Substitute back to find \( I_2 \)

Now substituting back:
\[ I_2 = 2 \left( \frac{\pi}{2} - 1 \right) = \pi - 2 \]

### Step 9: Substitute \( I_2 \) back to find \( I_4 \)

Now substituting \( I_2 \) back into the expression for \( I_4 \):
\[ I_4 = 4 \left( \frac{\pi^3}{8} - 3(\pi - 2) \right) \]

### Step 10: Calculate \( I_4 + 12I_2 \)

Finally, we can calculate \( I_4 + 12I_2 \):
\[ I_4 + 12I_2 = 4 \left( \frac{\pi^3}{8} - 3(\pi - 2) \right) + 12(\pi - 2) \]

After simplifying, we find:
\[ I_4 + 12I_2 = \frac{\pi^3}{2} - 12\pi + 48 \]

### Final Result

Thus, the final answer is:
\[ I_4 + 12I_2 = 4 \cdot \frac{\pi^3}{8} + 12(\pi - 2) = \frac{\pi^3}{2} + 12\pi - 24 \]