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int(sinax-sinbx)/(cosax-cosbx)dx |
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Answer» SOLUTION :`int(sinax-sinbx)/(cosax-cosbx)dx` =`(2SIN((a-b)/2) X.cos((a+b)/2) x)/(-2sin((a+b)/2) x.sin((a-b)/2) x)dx` =`-intcot((a+b)/2)xdx` =`-2/(a+b) In abs(sin((a+b)/2)x)+C` |
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