"(x + 2) dx = (x + 3) \\sin y. \\cos y dy"
"\\Rightarrow \\frac{x+2}{x+3} dx=\\frac{\\sin2y}{2}dy"
Integrating both sides, we get
"\\int\\frac{x+2}{x+3} dx=\\int\\frac{\\sin2y}{2}dy\n\\\\\\Rightarrow \\int\\frac{x+3-1}{x+3} dx=\\frac{1}{2}\\int\\sin2ydy\n\\\\\\Rightarrow \\int dx-\\int \\frac{1}{x+3}dx=-\\frac{1}{2}.\\frac{\\cos 2y}{2}+c\n\\\\\\Rightarrow x-ln|x+3|=-\\frac{\\cos 2y}{4}+c"
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