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RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\frac{3cos4x}{2sin2x}dx\]\[=\frac{3}{2}\int\frac{(12sin^22x)}{sin2x}dx\]\[=\frac{3}{2}\int\frac{1}{sin2x}\frac{2sin^22x}{sin2x}dx\]\[=\frac{3}{2}\int(\frac{1}{sin2x}2sin2x)dx\]

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.0note: \(cos4x =cos^2 2x  sin^2 2x\) \( \frac{3}{2} \int \frac{ \cos^2 2x  \sin^2 2x}{\sin 2x} dx\) =\( \frac{3}{2} [\int \frac{ \cos^2 2x}{\sin 2x} dx \int \sin 2x dx]\) =\( \frac{3}{2} [\int \cos^2 (2x)d(\cos 2x) \int \sin 2x dx]\) do you require further help?

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0@Shadowys The last step is wrong?

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0d(cos2x) gives sin2x instead of 1/sin2x
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