If m and n are integers such that (m - n) is | vedclass

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(B) We use the trigonometric identity $2 \sin(A) \cos(B) = \sin(A + B) + \sin(A - B)$.Here,$A = nx$ and $B = mx$,so $\cos(mx) \sin(nx) = \frac{1}{2} [

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$\frac{2n}{n^{2} - m^{2}}$

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(B) We use the trigonometric identity $2 \sin(A) \cos(B) = \sin(A + B) + \sin(A - B)$.Here,$A = nx$ and $B = mx$,so $\cos(mx) \sin(nx) = \frac{1}{2} [\sin((n + m)x) + \sin((n - m)x)]$.The integral becomes $I = \frac{1}{2} \int_{0}^{\pi} [\sin((n + m)x) + \sin((n - m)x)] \, dx$.Integrating,we get $I = \frac{1}{2} \left[ -\frac{\cos((n + m)x)}{n + m} - \frac{\cos((n - m)x)}{n - m} \right]_{0}^{\pi}$.Since $(m - n)$ is odd,$(m + n)$ is also odd. Thus,$\cos((n + m)\pi) = -1$ and $\cos((n - m)\pi) = -1$.Evaluating at the limits: $I = \frac{1}{2} \left[ (-\frac{-1}{n + m} - \frac{-1}{n - m}) - (-\frac{1}{n + m} - \frac{1}{n - m}) \right]$.$I = \frac{1}{2} \left[ \frac{1}{n + m} + \frac{1}{n - m} + \frac{1}{n + m} + \frac{1}{n - m} \right] = \frac{1}{n + m} + \frac{1}{n - m}$.$I = \frac{(n - m) + (n + m)}{n^{2} - m^{2}} = \frac{2n}{n^{2} - m^{2}}$.

If $m$ and $n$ are integers such that $(m - n)$ is an odd integer,then the value of the definite integral $\int_{0}^{\pi} \cos(mx) \sin(nx) \, dx$ is:

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