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dy/dx=sinxcosy how to find y?
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One way to do it is to get the dy on the same side as the y term and to get dx on the same side as the x term: \[\frac{dy}{\cos(y)}=\sin(x)dx\]From here, you can integrate both sides. The right side is fairly straightforward, and we get \[\int\limits{\frac{dy}{\cos(y)}}=\int\limits{\sin(x)dx}\]\[\int\limits{\frac{dy}{\cos(y)}}=-\cos(x)+K\] Now for the left side, the reciprocal of the cosine is the secant, so we have \[\int\limits{\sec(y)dy}=-\cos(x)+K\] Integrate (if you need help with this integral, click here: http://math2.org/math/integrals/more/sec.htm) \[\ln|\sec(y)+\tan(y)|=-\cos(x)+C\] Solve for y, if your professor/teacher requires it.
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