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Calculus: Integrals, antiderivatives, and Solving Differential Equations Symbolically:Separation of variables. y'=2xsecy solve for y
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\[\frac{dy}{sec{y}}=2x dx\] \[\int cos(y)dy=\int 2xdx\] \[sin {y} = x^2 +c\] \[y=sin^{-1}{x^2+c}\]
wait how do you get from step one to step two?
\[\frac{1}{sec{y}}=cos{y}\] It's a trig identity http://en.wikipedia.org/wiki/Trigonometric_functions
oh! okay thank you so much :))
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