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Integrate sin(nx)sin(x) over [0, pi] - Wave Equation
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multiply and divide the given expression by 2 then use the formula 2sin(A)sin(B) = cos(A - B ) - cos (A+B)
I end up with the answer \[1/2\left[ 1/(n+1)\sin (1+n)x - 1/(n-1)\sin (1-n)x \right]^{\pi}_{0}\] which according to me should be 0 for any value of n
what if n=1
\[\int\limits_{0}^{\pi}\sin(x)\sin(x)dx=\frac{\pi}{2}\] so it is not 0 everywhere
@zarkon for n= 1 it is pi/2 for the rest value of n it is coming out to be zero
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for positive integer n ...I know that
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