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 definition of the integral:
\[ I_{m,n} = \int \cos^m x \cos nx \, dx. \]

2. **Integration by Parts**:
We will use integration by parts. Let:
- \( u = \cos^m x \) (first part)
- \( dv = \cos nx \, dx \) (second part)

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

3. **Apply Integration by Parts**:
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_{m,n} = \left[ \cos^m x \cdot \frac{\sin nx}{n} \right] - \int \frac{\sin nx}{n} \cdot (-m \cos^{m-1} x \sin x) \, dx. \]
This simplifies to:
\[ I_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} \int \cos^{m-1} x \sin^2 x \cos nx \, dx. \]

4. **Use the Identity for \(\sin^2 x\)**:
Recall that \(\sin^2 x = 1 - \cos^2 x\). Thus, we can rewrite the integral:
\[ \int \cos^{m-1} x \sin^2 x \cos nx \, dx = \int \cos^{m-1} x (1 - \cos^2 x) \cos nx \, dx. \]
This expands to:
\[ \int \cos^{m-1} x \cos nx \, dx - \int \cos^{m+1} x \cos nx \, dx. \]
The first integral is \(I_{m-1,n}\) and the second integral can be rewritten as \(I_{m+1,n}\).

5. **Substituting Back**:
Substitute back into the equation:
\[ I_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} \left( I_{m-1,n} - I_{m+1,n} \right). \]

6. **Rearranging the Equation**:
Multiply through by \(n\) to eliminate the denominator:
\[ n I_{m,n} = \cos^m x \sin nx + m I_{m-1,n}. \]

7. **Final Rearrangement**:
Now, we can express \(I_{m,n}\) in terms of \(m+n\):
\[ (m+n) I_{m,n} = \cos^m x \sin nx + m I_{m-1,n-1}. \]

### Conclusion:
Thus, we have shown that

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