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OpenStudy (anonymous):
how would I integrate sin2x/2sinx dx
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OpenStudy (campbell_st):
well I would use the trig identity \[\sin(2x) = 2\sin(x)\cos(x) \] so the problem is now \[\frac{2\sin(x)\cos(x)}{2\sin(x)}\] then simplify the fraction... might make it a little easier.
OpenStudy (anonymous):
thank you
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