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Evaluate int(cos4x+cos2x)/(sin4x+sin2x)dx.

Answer»

SOLUTION :`INT(cos4x+cos2x)/(sin4x+sin2x)==int("2cos"(6x)/2"cos"(2X)/2)/(2"sin"(6x)/2"cos"(2x)/2)dx =intcot3xdx=1/3In|sin3x|+C`


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