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March 22nd, 2008, 10:22 AM   #1
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integral evaluation....

int (sin3x*sin5x)dx=?
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March 23rd, 2008, 10:55 AM   #2
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notice that: sinA*sinB=1/2[cos(A+B)-cos(A-B)] and cos2x= 1-2(sinx)^2

─▒ntegral becomes:

1/2 S[cos(8x)-cos(-2x)]dx = 1/2[sin8x/8 - sin2x/2] +C
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