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alfers101

  • 3 years ago

integral of 3cos4x / 2sin2x

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  1. RolyPoly
    • 3 years ago
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    \[\int\frac{3cos4x}{2sin2x}dx\]\[=\frac{3}{2}\int\frac{(1-2sin^22x)}{sin2x}dx\]\[=\frac{3}{2}\int\frac{1}{sin2x}-\frac{2sin^22x}{sin2x}dx\]\[=\frac{3}{2}\int(\frac{1}{sin2x}-2sin2x)dx\]

  2. Shadowys
    • 3 years ago
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    note: \(cos4x =cos^2 2x - sin^2 2x\) \( \frac{3}{2} \int \frac{ \cos^2 2x - \sin^2 2x}{\sin 2x} dx\) =\( \frac{3}{2} [\int \frac{ \cos^2 2x}{\sin 2x} dx- \int \sin 2x dx]\) =\( \frac{3}{2} [\int \cos^2 (2x)d(\cos 2x)- \int \sin 2x dx]\) do you require further help?

  3. RolyPoly
    • 3 years ago
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    @Shadowys The last step is wrong?

  4. Shadowys
    • 3 years ago
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    oh right. sorry.

  5. RolyPoly
    • 3 years ago
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    d(cos2x) gives -sin2x instead of 1/sin2x

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