How do you solve (secx)(dy/dx) = e^(y + sinx)?

dmitriylm
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Homework Statement


(secx)(dy/dx) = e^(y + sinx)


Homework Equations


None


The Attempt at a Solution


Not sure. I know I can do ln of both sides and isolate the y+sinx but then I get stuck with log^(secx)
 
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Ok, I got to (e^-y)dy = (e^sinx)(cosx)dx so I should be ok from there. Just posting in case someone is interested in the future.
 
Right. Writing the equation as sec(x) dy/dx= e^y e^{sin(x)} show that it is separable. e^{-y}dy= cos(x)e^{sin(x)}dx and both of those are easy to integrate.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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