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If $\mathrm{m}$ and $\mathrm{n}$ are integers, then what is the value of
$\int_{0}^{\pi} \sin m x \sin n x d x$, if $m \neq n ?$
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$\int_{0}^{\pi} \sin m x \sin n x d x$, if $m \neq n ?$
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Verified Answer
The correct answer is:
0
The given integral
$\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
$=\int_{0}^{\pi} \sin m(\pi-x) \cdot \sin n(\pi-x) d x$
$=\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
So, $\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
$=2 \int_{0}^{\pi / 2} \sin m x \cdot \sin n x d x$
$=2 \int_{0}^{\pi / 2} \frac{1}{2}[\cos (m x-n x)-\cos (m x+n x)] d x$
$=\int_{0}^{\pi / 2}[\cos (m-n) x-\cos (m+n) x] d x$
$=\left[\frac{\sin (m-n) x}{m-n}-\frac{\sin (m+n) x}{m+n}\right]_{0}^{\pi / 2}=0$
$\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
$=\int_{0}^{\pi} \sin m(\pi-x) \cdot \sin n(\pi-x) d x$
$=\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
So, $\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
$=2 \int_{0}^{\pi / 2} \sin m x \cdot \sin n x d x$
$=2 \int_{0}^{\pi / 2} \frac{1}{2}[\cos (m x-n x)-\cos (m x+n x)] d x$
$=\int_{0}^{\pi / 2}[\cos (m-n) x-\cos (m+n) x] d x$
$=\left[\frac{\sin (m-n) x}{m-n}-\frac{\sin (m+n) x}{m+n}\right]_{0}^{\pi / 2}=0$
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