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  • 4 years ago

show that for all integers "m" and "n", with "m" is not equal to positive or negative "n", the integral of cos(m*x)*cos(n*x) dx in the interval (pi,-pi) equals zero.

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  1. anonymous
    • 4 years ago
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    \[\int\limits \cos(mx) \cos(nx) dx\] = \[1/2\int\limits [\cos{x (m-n)}+\cos{x (m+n)}] dx \] = \[1/2\int\limits [\cos(m x-n x)+\cos(m x+n x)] dx\] = \[1/2 \int\limits \cos(m x-n x) dx+1/2 \int\limits \cos(m x+n x) dx\] = \[ 1/2 \int\limits \cos(x (m-n)) dx+1/2 \int\limits \cos(m x+n x) dx\] For the integrand cos(x (m-n)), substitute u = x (m-n) and du = m-n dx: = \[ 1/[2 (m-n)] \int\limits \cos(u) du+1/2 \int\limits \cos(m x+n x) dx\] = \[ \sin(u)/[2 (m-n)]+1/2 \int\limits \cos(m x+n x) dx\] = \[ \sin(u)/[2 (m-n)]+1/2 \int\limits \cos(x (m+n)) dx\] For the integrand cos(x (m+n)), substitute s = x (m+n) and ds = m+n dx: = \[ 1/[2 (m+n)] \int\limits \cos(s) ds+\sin(u)/[2 (m-n)]\] = sin(s)/[2 (m+n)]+sin(u)/[2 (m-n)]+constant Substitute back for s = x (m+n) and u = x (m-n) to get the integral as : (m sin(m x) cos(n x)-n cos(m x) sin(n x))/(m^2-n^2)+constant Now you can work on with the intervals part.

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