the question is from equation of first order and first degree

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Divide by 1+2sin y+cos y.

`[(3+2sin x+ cos x)dy]/(1+2sin y+cos y)=dx`

Divide by 3+2sin x+ cos x.

`(dy)/(1+2sin y+cos y)=(dx)/(3+2sin x+ cos x)`

Use integrals to calculate`(dy)/(1+2sin y+cos y) and (dx)/(3+2sin x+ cos x).`

Use the formula `sin y=2 tan(y/2)/(1+tan^2(y/2))` `cos y =(1- tan^2(y/2))/(1+tan^2(y/2))`

Substitute `tan (y/2)=u`

`` `(1+tan^2(y/2))/2dy=du=gtdy=(2du)/(1+u^2)` `int (dy)/(1+2sin y+cos y)=int [(2du)/(1+u^2)]/(1+4u/(1+u^2)+(1-u^2)/(1+u^2))`

`int (2du)/(1+u^2+4u+1-u^2)=int (2du)/(2+4u) = int (du)/(1+2u)`

Substitute `1+2u=v =gt 2du=dv =gt du=(dv)/2` `int (du)/(1+2u) = int (dv)/2v = (1/2)*ln v + c= (ln(1+2u))/2 + c= (ln (1+2 tan(y/2))/2 + c`

`int (dy)/(1+2sin y+cos y) =ln (1+2 tan(y/2))/2 + c`

Calculate integral of  `(dx)/(3+2sin x+ cos x).`

Substitute `tan (x/2)=p =gt dx=(2dp)/(1+p^2)` `int (dx)/(3+2sin x+ cos x) = int ((2dp)/(1+p^2))/(3 + 4p/(1+p^2)+(1-p^2)/(1+p^2))` `int (2dp)/(3+3p^2+4p+1-p^2) = int (2dp)/(2p^2+4p+4) = int (dp)/(p^2+2p+2)`

Create arctan function

`int (dp)/((p^2+2p+1)-1+2) =int (dp)/((p+1)^2+1) = arctan (p+1)+c`

`` `int dx/(3+2sin x+ cos x) = arctan (tan (x/2)+1) + c`

ANSWER: `ln (1+2 tan(y/2))/2 = arctan (tan (x/2)+1) + c`

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