Given,
cosy⋅dxdy=sin2x×cosx cosy×dy=sin2x×cosxdx
Now, Integrating both sides
∫cosydy=∫sin2xcosxdx
Let I1=∫sin2xcosxdx
Substitute sinx = t
then cosxdx = dt
I1=∫t2dtI1=3t3
Now, I1=31sin3x
So,
∫cosydy=∫sin2xcosxdx
siny=31sin3x+Cy=sin−1(31sin3x+C)
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