Solve int y wrt x + 2 x d y = ∫ (from 0 to 2 pi) of (1-cost)(t-sint)^prime+2(t-sint)(1-cost)^primedt

$$\int y d x + 2 x d y = \int _ { 0 } ^ { 2 \pi } ( 1 - \cos t ) ( t - \sin t ) ^ { \prime } + 2 ( t - \sin t ) ( 1 - \cos t ) ^ { \prime } d t$$

Solve for d

$$\left\{\begin{matrix}d=-\frac{1}{2}+\frac{С}{xy}\text{, }&y\neq 0\text{ and }x\neq 0\\d\in \mathrm{R}\text{, }&\left(x=0\text{ or }y=0\right)\text{ and }С=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract \int y\mathrm{d}x from both sides.

$$2xdy=\int _{0}^{2\pi }\left(1-\cos(t)\right)\frac{\mathrm{d}}{\mathrm{d}x}(t-\sin(t))+2\left(t-\sin(t)\right)\frac{\mathrm{d}}{\mathrm{d}x}(1-\cos(t))\mathrm{d}t-\int y\mathrm{d}x$$

Use the distributive property to multiply 1-\cos(t) by \frac{\mathrm{d}}{\mathrm{d}x}(t-\sin(t)).

$$2xdy=\int _{0}^{2\pi }\frac{\mathrm{d}}{\mathrm{d}x}(t-\sin(t))-\cos(t)\frac{\mathrm{d}}{\mathrm{d}x}(t-\sin(t))+2\left(t-\sin(t)\right)\frac{\mathrm{d}}{\mathrm{d}x}(1-\cos(t))\mathrm{d}t-\int y\mathrm{d}x$$

Use the distributive property to multiply 2 by t-\sin(t).

$$2xdy=\int _{0}^{2\pi }\frac{\mathrm{d}}{\mathrm{d}x}(t-\sin(t))-\cos(t)\frac{\mathrm{d}}{\mathrm{d}x}(t-\sin(t))+\left(2t-2\sin(t)\right)\frac{\mathrm{d}}{\mathrm{d}x}(1-\cos(t))\mathrm{d}t-\int y\mathrm{d}x$$

Use the distributive property to multiply 2t-2\sin(t) by \frac{\mathrm{d}}{\mathrm{d}x}(1-\cos(t)).

$$2xdy=\int _{0}^{2\pi }\frac{\mathrm{d}}{\mathrm{d}x}(t-\sin(t))-\cos(t)\frac{\mathrm{d}}{\mathrm{d}x}(t-\sin(t))+2t\frac{\mathrm{d}}{\mathrm{d}x}(1-\cos(t))-2\sin(t)\frac{\mathrm{d}}{\mathrm{d}x}(1-\cos(t))\mathrm{d}t-\int y\mathrm{d}x$$

The equation is in standard form.

$$2xyd=-xy-С$$

Divide both sides by 2xy.

$$\frac{2xyd}{2xy}=\frac{-xy-С}{2xy}$$

Dividing by 2xy undoes the multiplication by 2xy.

$$d=\frac{-xy-С}{2xy}$$

Divide -yx-С by 2xy.

$$d=-\frac{1}{2}+\frac{С}{xy}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{С}{y\left(2d+1\right)}\text{, }&d\neq -\frac{1}{2}\text{ and }y\neq 0\\x\in \mathrm{R}\text{, }&\left(y=0\text{ or }d=-\frac{1}{2}\right)\text{ and }С=0\end{matrix}\right.$$