- #1
JoshHolloway
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[tex]y \prime = (cos^{2}x)(cos^{2}(2y))[/tex]
[tex] \frac{dy}{dx} = \frac{1}{2} (cox(2x) + 1) \frac{1}{4}(cos(4y) + 1)[/tex]
[tex] \frac{1}{cos(4y) + 1}dy \ = \ \frac{1}{8} (cos(2x) + 1)dx [/tex]
[tex] \int \frac{1}{ \sqrt{cos(4y)}^{2} + 1}dy \ = \ \frac{1}{8} \int (cos(2x) + 1)dx [/tex]
[tex] tan^{-1}(cos4y) \ = \ \frac{1}{8} \frac{1}{2} sin(2x) + \frac{1}{8} x + C [/tex]
[tex] tan[tan^{-1}(cos4y) \ = \ \frac{1}{16} sin(2x) + \frac{1}{8} x + C] [/tex]
[tex] cos^{-1}[cos(4y) \ = \ tan( \frac{sin2x}{16} + \frac{x}{8} + C)] [/tex]
[tex] y \ = \ \frac{1}{4} cos^{-1}[tan( \frac{sin2x}{16} + \frac{x}{8} + C)] [/tex]
[tex] \frac{dy}{dx} = \frac{1}{2} (cox(2x) + 1) \frac{1}{4}(cos(4y) + 1)[/tex]
[tex] \frac{1}{cos(4y) + 1}dy \ = \ \frac{1}{8} (cos(2x) + 1)dx [/tex]
[tex] \int \frac{1}{ \sqrt{cos(4y)}^{2} + 1}dy \ = \ \frac{1}{8} \int (cos(2x) + 1)dx [/tex]
[tex] tan^{-1}(cos4y) \ = \ \frac{1}{8} \frac{1}{2} sin(2x) + \frac{1}{8} x + C [/tex]
[tex] tan[tan^{-1}(cos4y) \ = \ \frac{1}{16} sin(2x) + \frac{1}{8} x + C] [/tex]
[tex] cos^{-1}[cos(4y) \ = \ tan( \frac{sin2x}{16} + \frac{x}{8} + C)] [/tex]
[tex] y \ = \ \frac{1}{4} cos^{-1}[tan( \frac{sin2x}{16} + \frac{x}{8} + C)] [/tex]
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