calculus

Find the general solution of this de (1+cosx)y'=sinx(sinx+sinxcosx-y)

I need full step have tried but nothing good is coming out

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  1. Rearrange things into the usual form for a 1st-order linear DE:
    (1+cosx)y' = sin^2x(1+cosx) - sinx y
    y' - sinx/(1+cosx) y = sin^2x
    y' - tan(x/2) y = sin^2x
    Now you have the form y' + p(x) y = q(x)
    The integrating factor is now e^∫p(x) dx = 1/2 sec^2(x/2)
    1/2 sec^2(x/2) y' - tan(x/2)(1/2 sec^2(x/2))y = sin^2(x) * 1/2 sec^2(x/2)
    After some trig manipulation, this comes out to be
    (tan(x/2) y)' = 4(1+cos^2(x))/(1+cos(x))
    Some more study should provide the way to integrate the right side.

    Check my math, and don't be afraid to use tools like wolframalpha to provide some help.

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    oobleck
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