How to solve differential equation $\frac{d}{dx}\left(\frac{\lambda y'}{\sqrt{1+y'^2}}\right)=1$ My task is to solve for $y$ from:
$$\frac{d}{dx}\left(\frac{\lambda y'}{\sqrt{1+y'^2}}\right)=1$$
I have been given the answer, but I would like to calculate this myself also. $\lambda$ is a constant. How should I proceed? 
The answer according to source material: 

Integrating with respect to $x$ we get:
$$x+C_1= \lambda\sin(\theta)$$
$$y=-\lambda\cos(\theta)+C_2$$

 A: hint:
integrating once you get $$\dfrac{y^\prime}{\sqrt{1+{y^\prime}^2}} = \dfrac{x + C}{\lambda}  \tag 1$$
(a) square $(1)$ and find $y^\prime$
(b) solve the first order equation.

edit:
since the left hand side of $(1)$ is smaller than one in absolute value, it can be set to $\sin \theta.$ which gives 
$$x + C = \sin \theta, \  \dfrac{y^\prime}{\sqrt{1+{y^\prime}^2}} = \sin \theta$$
solving the second equation  gives $y^\prime = \tan \theta.$ therefore 
$$\frac{dy}{d \theta} = \frac{dy}{dx} \frac{dx}{d \theta} = \tan \theta \cos \theta = \sin \theta$$ which has the solution $$y = \cos \theta + D$$
A: Here is my own try: 
$$\frac{\lambda y'}{\sqrt{1+y'^2}}=x+C$$
$$\frac{ y'}{\sqrt{1+y'^2}}=\frac{x+C}{\lambda}$$
$$\frac{y'^2}{1+y'^2}=\frac{\left(x+C\right)^2}{\lambda^2}$$
$$\frac{-1}{1+y'^2}=\frac{\left(x+C\right)^2-\lambda^2}{\lambda^2}$$
$$1+y'^2=\frac{-\lambda^2}{\left(x+C\right)^2-\lambda^2}$$
$$y'^2=\frac{-\left(x+C\right)^2}{\left(x+C\right)^2-\lambda^2}$$
$$y'^2=\frac{\left(x+C\right)^2}{\lambda^2-\left(x+C\right)^2}$$
$$\int \;dy=\pm\int \frac{x+C}{\sqrt{\lambda^2-\left(x+C\right)^2}}\;dx$$
$$u=x+C,\;du=dx$$
$$y = \int \;dy=\pm \int \frac{u}{\sqrt{\lambda^2-u^2}}\;du = \pm \;-\sqrt{\lambda^2-u^2}+D=\pm\sqrt{\lambda^2-\left(x+C\right)^2}+D$$
so
$$y(x)=\pm \sqrt{\lambda^2-\left(x+C\right)^2}+D$$
