If `I_(n)=int_(0)^(pi)x^(n)sinxdx`, then find the value of `I_(5)+20I_(3)`.

Explanation

To solve the problem, we need to evaluate the expression \( I_5 + 20I_3 \) where \( I_n = \int_0^{\pi} x^n \sin x \, dx \).

### Step-by-Step Solution:

1. **Define the integral**:
\[ I_n = \int_0^{\pi} x^n \sin x \, dx \]

2. **Use integration by parts**:
We will apply integration by parts where we let:
- \( u = x^n \) and \( dv = \sin x \, dx \)
- Then, \( du = n x^{n-1} \, dx \) and \( v = -\cos x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_n = \left[-x^n \cos x \right]_0^{\pi} + \int_0^{\pi} n x^{n-1} \cos x \, dx \]

3. **Evaluate the boundary term**:
- At \( x = \pi \): \( -\pi^n \cos(\pi) = \pi^n \)
- At \( x = 0 \): \( -0^n \cos(0) = 0 \)
Thus:
\[ I_n = \pi^n + n \int_0^{\pi} x^{n-1} \cos x \, dx \]

4. **Define the new integral**:
Let \( J_n = \int_0^{\pi} x^{n-1} \cos x \, dx \). Then we have:
\[ I_n = \pi^n + n J_{n-1} \]

5. **Apply integration by parts to \( J_n \)**:
For \( J_n \), we apply integration by parts again:
- Let \( u = x^{n-1} \) and \( dv = \cos x \, dx \)
- Then \( du = (n-1)x^{n-2} \, dx \) and \( v = \sin x \)

Thus:
\[ J_n = \left[x^{n-1} \sin x \right]_0^{\pi} - \int_0^{\pi} (n-1)x^{n-2} \sin x \, dx \]

6. **Evaluate the boundary term for \( J_n \)**:
- At \( x = \pi \): \( \pi^{n-1} \sin(\pi) = 0 \)
- At \( x = 0 \): \( 0^{n-1} \sin(0) = 0 \)
Thus:
\[ J_n = -(n-1) I_{n-2} \]

7. **Substituting back**:
Now substituting \( J_{n-1} \) into the equation for \( I_n \):
\[ I_n = \pi^n + n \left[-(n-1) I_{n-2}\right] \]
Simplifying gives:
\[ I_n = \pi^n - n(n-1) I_{n-2} \]

8. **Finding \( I_5 \) and \( I_3 \)**:
- For \( n = 5 \):
\[ I_5 = \pi^5 - 5 \cdot 4 I_3 \]
- For \( n = 3 \):
\[ I_3 = \pi^3 - 3 \cdot 2 I_1 \]

9. **Finding \( I_1 \)**:
\[ I_1 = \int_0^{\pi} x \sin x \, dx \]
Using integration by parts:
\[ I_1 = \left[-x \cos x \right]_0^{\pi} + \int_0^{\pi} \cos x \, dx \]
Evaluating gives:
\[ I_1 = \pi + 0 = \pi \]

10. **Substituting \( I_1 \) back into \( I_3 \)**:
\[ I_3 = \pi^3 - 6\pi = \pi^3 - 6\pi \]

11. **Substituting \( I_3 \) back into \( I_5 \)**:
\[ I_5 = \pi^5 - 20(\pi^3 - 6\pi) = \pi^5 - 20\pi^3 + 120\pi \]

12. **Calculate \( I_5 + 20I_3 \)**:
\[ I_5 + 20I_3 = \left(\pi^5 - 20\pi^3 + 120\pi\right) + 20\left(\pi^3 - 6\pi\right) \]
Simplifying gives:
\[ I_5 + 20I_3 = \pi^5 - 20\pi^3 + 120\pi + 20\pi^3 - 120\pi = \pi^5 \]

### Final Answer:
\[ I_5 + 20I_3 = \pi^5 \]