Derive reduction formula for
`l_((n","m))=int (sin^(n)x)/(cos^(m)x)dx`.

Answer

To derive the reduction formula for \( I(n, m) = \int \frac{\sin^n x}{\cos^m x} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
We start with the integral:
\[ I(n, m) = \int \frac{\sin^n x}{\cos^m x} \, dx \]
We can express \(\sin^n x\) as \(\sin^{n-1} x \cdot \sin x\):
\[ I(n, m) = \int \frac{\sin^{n-1} x \cdot \sin x}{\cos^m x} \, dx \]

2. **Integration by Parts**:
We will use integration by parts. Let:
- \( u = \sin^{n-1} x \)
- \( dv = \frac{\sin x}{\cos^m x} \, dx \)

Then, we need to find \( du \) and \( v \):
- \( du = (n-1) \sin^{n-2} x \cos x \, dx \)
- To find \( v \), we can integrate \( dv \):
\[ v = \int \frac{\sin x}{\cos^m x} \, dx \]

3. **Applying Integration by Parts**:
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ I(n, m) = \sin^{n-1} x \cdot v - \int v \cdot du \]

4. **Evaluate \( v \)**:
The integral \( v = \int \frac{\sin x}{\cos^m x} \, dx \) can be evaluated using substitution. Let \( u = \cos x \), then \( du = -\sin x \, dx \):
\[ v = -\int \frac{1}{u^m} \, du = -\frac{u^{-m+1}}{-m+1} = \frac{\cos^{1-m} x}{1-m} \]

5. **Substituting Back**:
Now substituting \( v \) back into our integration by parts result:
\[ I(n, m) = \sin^{n-1} x \cdot \frac{\cos^{1-m} x}{1-m} - \int \frac{\cos^{1-m} x}{1-m} \cdot (n-1) \sin^{n-2} x \cos x \, dx \]
This gives:
\[ I(n, m) = \frac{\sin^{n-1} x \cos^{1-m} x}{1-m} - \frac{(n-1)}{1-m} I(n-2, m) \]

6. **Final Reduction Formula**:
Rearranging gives us the reduction formula:
\[ I(n, m) = \frac{\sin^{n-1} x \cos^{1-m} x}{1-m} - \frac{(n-1)}{1-m} I(n-2, m) \]