Answer :

Given



This is a first order linear differential equation of the form



Here, P = cos x and Q = sin x cos x


The integrating factor (I.F) of this differential equation is,



We have


I.F = esin x


Hence, the solution of the differential equation is,





Let sin x = t


cosxdx = dt [Differentiating both sides]


By substituting this in the above integral, we get




Recall





yet = tet – et + c


yet × e–t = (tet – et + c)e–t


y = t – 1 + ce–t


y = sin x – 1 + ce–sin x [ t = sin x]


Thus, the solution of the given differential equation is y = sin x – 1 + ce–sin x


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