dy/dx =[ (e^y)(sin x)^2 ] / [y (sec x)] sorry about the mess, i'm still learning how to use latex
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Originally Posted by crafty dy/dx =[ (e^y)(sin x)^2 ] / [y (sec x)] sorry about the mess, i'm still learning how to use latex Your equation separates $\displaystyle y e^{-y}\,dy = \sin^2x \cos x \, dx$ so $\displaystyle \int y e^{-y}\,dy = \int \sin^2x \cos x \, dx + c$ Just need to carry out the integrations.
(e^(-y))(-y - 1) = (1/3)(sin x)^3 + C this is what i get but do i have to put this in terms of y and if so how would i do that?
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