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Solve DE: csc(y)dx=sec^2(x)dy
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\[\frac{1}{\sin(y)}dx=\frac{1}{\cos^2(x)}dy\]
\[\cos^2(x)dx=\sin(y)dy\] i'm guessing double angle theorem?
the right side is pretty simple
\[\int\limits_{}{}\sin(y)dy=-\cos(y)+c_1\]
going to have to refresh myself on double angles
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\[\int\limits_{}{}\cos^2(x)=\int\limits{}{}\frac{1+\cos(2x)}{2}=\frac{1}{2}\int\limits{}{}dx+\frac{1}{2}\int\limits{}{}\cos(2x)dx\]
\[=\frac{1}{2}x+\frac{1}{4}\sin(2x)+c_2\]
other than forgetting a negative -(-cos(y)=cos(y) it's right =]
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