If `I_(m"," n)=int cos^(m)x*cos nx dx`, show that `(m+n)I_(m","n)=cos^(m)x*sin nx+m I_((m-1","n-1))`

Answer

To solve the problem, we need to show that

\[ (m+n)I_{m,n} = \cos^m x \sin nx + m I_{m-1,n-1} \]

where

\[ I_{m,n} = \int \cos^m x \cos nx \, dx. \]

### Step-by-Step Solution:

1. **Define the Integral**:
We start with the integral definition:
\[ I_{m,n} = \int \cos^m x \cos nx \, dx. \]

2. **Apply Integration by Parts**:
We can use integration by parts, where we let:
- \( u = \cos^m x \) (which we will differentiate)
- \( dv = \cos nx \, dx \) (which we will integrate)

Then, we differentiate \( u \) and integrate \( dv \):
- \( du = -m \cos^{m-1} x \sin x \, dx \)
- \( v = \frac{\sin nx}{n} \)

Applying integration by parts:
\[ I_{m,n} = uv - \int v \, du \]
becomes:
\[ I_{m,n} = \cos^m x \cdot \frac{\sin nx}{n} - \int \frac{\sin nx}{n} \cdot (-m \cos^{m-1} x \sin x) \, dx. \]

3. **Simplify the Integral**:
The integral simplifies to:
\[ I_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} \int \cos^{m-1} x \sin nx \sin x \, dx. \]

4. **Use the Product-to-Sum Formula**:
We can use the product-to-sum identities to simplify \(\sin nx \sin x\):
\[ \sin nx \sin x = \frac{1}{2} [\cos(n-1)x - \cos(n+1)x]. \]

Thus, we can rewrite the integral:
\[ I_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{2n} \left( \int \cos^{m-1} x \cos(n-1)x \, dx - \int \cos^{m-1} x \cos(n+1)x \, dx \right). \]

5. **Identify the New Integrals**:
The integrals we have now can be represented as:
\[ I_{m-1,n-1} = \int \cos^{m-1} x \cos(n-1)x \, dx, \]
\[ I_{m-1,n+1} = \int \cos^{m-1} x \cos(n+1)x \, dx. \]

6. **Combine the Results**:
Now we can express \(I_{m,n}\) in terms of \(I_{m-1,n-1}\) and \(I_{m-1,n+1}\):
\[ I_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{2n} (I_{m-1,n-1} - I_{m-1,n+1}). \]

7. **Multiply by \(n\)**:
Multiply the entire equation by \(n\):
\[ n I_{m,n} = \cos^m x \sin nx + \frac{m}{2} (I_{m-1,n-1} - I_{m-1,n+1}). \]

8. **Rearranging**:
Rearranging gives us:
\[ (m+n) I_{m,n} = \cos^m x \sin nx + m I_{m-1,n-1}. \]

### Conclusion:
Thus, we have shown that:

\[ (m+n) I_{m,n} = \cos^m x \sin nx + m I_{m-1,n-1}. \]