Can you find general solution of this equation dy/dx= (1+cosx)/(1-cosx).

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int y *dy = int (1+cosx)/(1-cosx) *dx int y *dy= int ( 2 cos^2 x/2) / (2 sin^2 x/2) *dx int y *dy= int (cos^x/2) / (sin^2 x/2) * dx int y *dy= int (cot^2 x/2) * dx y = - (cosec ^ 2 x/2 ) / (1/2) +c y = -2 cosec ^2 x/2 ) +c
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Maths Tutor

y=-2Cot(x/2)-x+Constant
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-2cot(x) - 2x
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Trainer

dy/dx= (1+2cos^2(x/2)-1)/(1-(1-2sin^2(x/2))= Cot^2(x/2) dy/dx= 1-cosec^2(x/2) Int(dy)= Int(1-cosec^2(x/2))dx y = x- 2 cot(x/2) + C
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Physics Tutor

dy/dx=(1+cosx)/(1- cosx) => dy/dx=(2cos^2(x/2))/(2sin^2(x/2)) =>dy/dx=cot^2(x/2) => dy/dx=csc^2(x/2)-1 =>Integral=Integral =>y=2cot(x/2) - x + k
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Physics coach

It is formula based question.
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Semi Qualified Chartered Accountant

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Maths, Organic Chemistry and Physics Tutor

Multiply the numerator and denominator with (1+cos x)..nd then proceed.
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Tutor

Express 1+cosx and 1-cosx in half multiple angles and eqn reduces to dy = (cot(x/2))^2*dx Just integrate it
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B.Tech

For that you have to multiply 1 + cosx in the numerator and denominator then write the value of (1+cosx)^2 = 1 + (cosx)^2 + 2cosx and divide that value with 1 - (cosx)^2 = (sinx)^2 So you will get (cosecx)^2 + (cotx)^2 + 2 cotx .cosecx now integrate wrt to x and u will get the value y = 2cotx -cosecx-x...
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