Why Does \int^{1}_{0}sin(\pi x)sin(n\pi x)dx = 1 for n = 1 and 0 for n >= 2?

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SUMMARY

The integral \(\int^{1}_{0} \sin(\pi x) \sin(n \pi x) dx\) evaluates to 1 when \(n = 1\) and 0 for all integers \(n \geq 2\). This result is derived from the orthogonality property of sine functions over the interval [0, 1]. Specifically, the trigonometric addition formulas facilitate the simplification of the integral, confirming the stated outcomes based on the value of \(n\).

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Can somebody explain to me why
\int^{1}_{0}sin( \pi x) sin (n \pi x) dx equals 1 if n = 1 and 0 for n >= 2? (n is an integer)
 
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Just to give a hint: trigonometric addition formulas...
If this isn't sufficient to solve it, ask again.
 

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