`int_(0)^(pi) sin^(n) x.cos^(2m+1) xdx` is equal to

Options

A. 0
B. `int_(0)^(pi) cos^(2m-1) xdx`
C. `((2m+1)!)/(n!)`
D. None of these

Answer

To solve the integral \( I = \int_{0}^{\pi} \sin^n x \cos^{2m+1} x \, dx \), we can use a symmetry property of definite integrals. Here’s a step-by-step solution:

### Step 1: Define the integral
Let
\[ I = \int_{0}^{\pi} \sin^n x \cos^{2m+1} x \, dx \]

### Step 2: Use the property of definite integrals
We can use the property of integrals that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we set \( a = \pi \). Thus, we have:
\[ I = \int_{0}^{\pi} \sin^n(\pi - x) \cos^{2m+1}(\pi - x) \, dx \]

### Step 3: Simplify the integrand
Using the identities \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:
\[ I = \int_{0}^{\pi} \sin^n x \cdot (-\cos(2m+1)x) \, dx \]
This simplifies to:
\[ I = -\int_{0}^{\pi} \sin^n x \cos^{2m+1} x \, dx \]

### Step 4: Relate the two integrals
Now we have:
\[ I = -I \]
Adding \( I \) to both sides gives:
\[ 2I = 0 \]

### Step 5: Solve for \( I \)
Dividing both sides by 2 yields:
\[ I = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \int_{0}^{\pi} \sin^n x \cos^{2m+1} x \, dx = 0 \]