If m, n are integers and m + n is odd then the value of $$∫_0^πsin⁡mx.cos⁡nx.dx$$​ is

Correct option is A

​$$\text{Given,} \\I = \int_0^{\pi} \sin(mx) \cdot \cos(nx) \, dx \\[8pt]\text{We know that} \\\sin A \cdot \cos B = \frac{1}{2} \left[ \sin(A + B) + \sin(A - B) \right] \\[8pt]I = \int_0^{\pi} \frac{1}{2} \left[ \sin(mx + nx) + \sin(mx - nx) \right] dx \\[8pt]I = \frac{1}{2} \int_0^{\pi} \left[ \sin((m + n)x) + \sin((m - n)x) \right] dx \\[8pt]I = \frac{1}{2} \int_0^{\pi} \sin((m + n)x) \, dx + \frac{1}{2} \int_0^{\pi} \sin((m - n)x) \, dx \\[8pt]I = -\frac{1}{2} \left[ \frac{\cos((m - n)x)}{m - n} \right]_0^{\pi} + \frac{1}{2} \left[ \frac{\cos((m + n)x)}{m + n} \right]_0^{\pi} \\[8pt]I = -\frac{1}{2} \left[ \frac{\cos((m - n)\pi)}{m - n} - \frac{\cos((m - n) \cdot 0)}{m - n} \right] + \frac{1}{2} \left[ \frac{\cos((m + n)\pi)}{m + n} - \frac{\cos((m + n) \cdot 0)}{m + n} \right] \\[8pt]I = 0$$​​