If (n - m) is odd and |m| ne |n|, then int_0^ | vedclass

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(A) Let $I = \int_0^\pi \cos mx \sin nx \,dx$. Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have $\cos mx \sin nx = \

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

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(A) Let $I = \int_0^\pi \cos mx \sin nx \,dx$. Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have $\cos mx \sin nx = \frac{1}{2} [\sin(n+m)x - \sin(m-n)x] = \frac{1}{2} [\sin(n+m)x + \sin(n-m)x]$. Integrating this,we get: $I = \frac{1}{2} \int_0^\pi [\sin(n+m)x + \sin(n-m)x] \,dx$ $I = \frac{1}{2} \left[ -\frac{\cos(n+m)x}{n+m} - \frac{\cos(n-m)x}{n-m} \right]_0^\pi$ $I = -\frac{1}{2} \left[ \left( \frac{\cos(n+m)\pi}{n+m} + \frac{\cos(n-m)\pi}{n-m} \right) - \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right]$ Since $(n-m)$ is odd,$(n+m)$ is also odd. Thus,$\cos(n+m)\pi = -1$ and $\cos(n-m)\pi = -1$. $I = -\frac{1}{2} \left[ \left( \frac{-1}{n+m} - \frac{1}{n-m} \right) - \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right]$ $I = -\frac{1}{2} \left[ -2 \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right] = \frac{1}{n+m} + \frac{1}{n-m}$ $I = \frac{(n-m) + (n+m)}{n^2 - m^2} = \frac{2n}{n^2 - m^2}$.

If $(n - m)$ is odd and $|m| \ne |n|,$ then $\int_0^\pi {\cos mx \sin nx} \,dx$ is

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