Ordinary differential equation $y'(t)=\sin(f(t,y))$ One whose solution never makes me happy is the following:
$$y'(t)=\sin(y+t)\text{.}$$
I would start by substituting $z(t)=y(t)+t$ to get an ODE in $z(t)$, but then I'm not sure about how to substitute back my solution to check if it's correct or not$\dots$
 A: All right, now that you've got a solution, let's test it.  Starting with the right side, we have
$$\sin(y+t)=\sin[2\tan^{-1}(-1-\frac2{t+c})]=2\sin[\tan^{-1}(-1-\frac2{t+c})]\cos[\tan^{-1}(-1-\frac2{t+c})]=$$
$$2\tan[\tan^{-1}(-1-\frac2{t+c})]\cos^2[\tan^{-1}(-1-\frac2{t+c})]=$$
$$\dfrac{2\tan[\tan^{-1}(-1-\frac2{t+c})]}{1+\tan^2[\tan^{-1}(-1-\frac2{t+c})]}=\dfrac{-2-\frac4{t+c}}{1+(-1-\frac2{t+c})^2}=-\dfrac{2+\frac4{t+c}}{2+\frac4{t+c}+\frac4{(t+c)^2}}$$
Now let's work with the left side
$$y'=\frac d{dt}[2\tan^{-1}(-1-\frac2{t+c})-t]=\dfrac{\frac4{(t+c)^2}}{1+(-1-\frac2{t+c})^2}-\dfrac{1+(-1-\frac2{t+c})^2}{1+(-1-\frac2{t+c})^2}=$$
$$\dfrac{\frac4{(t+c)^2}-2-\frac4{t+c}-\frac4{(t+c)^2}}{2+\frac4{t+c}+\frac4{(t+c)^2}}=-\dfrac{2+\frac4{t+c}}{2+\frac4{t+c}+\frac4{(t+c)^2}}$$
It was a lot of work, but it appears your answer checks.
A: Your proposed substitution transforms the ODE to
$$z'(t)-1=\sin z(t), $$
which can be solved by separation of variables:
$$\int \frac{\mathrm{d}z}{1+\sin z}=\int \mathrm{d}t.$$
You should get $$\frac{2}{\cot (z/2)+1}=t+c, $$
which simplifies to 
$$z= 2 \text{ arccot } \left(\frac{2}{t+c}-1 \right). $$
Finally since $y=z-t$ the general solution to the original equation is
$$ y=2 \text{ arccot } \left(\frac{2}{t+c}-1 \right)-t.$$
A: (Thanks to Giuseppe Negro)
A.D. Polyanin and  V.F. Zaitsev, in Handbook of Exact Solutions for Ordinary Differential Equations, 2nd ed. 2003, state (equation 20, p. 201):

$y'_x=f(x)\sin(\lambda y+g(x))$.
The substituion $w=\lambda y+g(x)$ leads to an equation of the form [$\ldots$]: $w'_x=\lambda f(x)\sin(w)+g'_x(x)$.

The latter may be further reduced to a Riccati equation (see equation 11, p. 201), but in this case $f(x)\equiv 1$, $\lambda=1$ and $g(x)=\mathrm{Id}(x)$, the equation is separable and integration leads to the solution
$${y=2\text{arccot}\left(\frac{2}{t+c}-1\right)-t}\ ,\quad{c\in\mathbb{R}}$$
already agreed.

Addendum:
Last but not least, thanks to Mike:
$$\sin(y+t)=
\sin\left[2\cot^{-1}\left(\dfrac2{t+c}-1\right)\right]=
2\sin\left[\cot^{-1}\left(\dfrac2{t+c}-1\right)\right]\cos\left[\cot^{-1}\left(\dfrac2{t+c}-1\right)\right]$$
$$=2\sin^2\left[\cot^{-1}\left(\dfrac2{t+c}-1\right)\right]\cot\left[\cot^{-1}\left(\dfrac2{t+c}-1\right)\right]=$$
$$=\dfrac{2\cot\left[\cot^{-1}\left(\dfrac2{t+c}-1\right)\right]}{1+\cot^2\left[\cot^{-1}\left(\dfrac2{t+c}-1\right)\right]}=
\dfrac{-2+\dfrac4{t+c}}{1+\left(\dfrac2{t+c}-1\right)^2}=
\dfrac{-2+\dfrac4{t+c}}{2-\dfrac4{t+c}+\dfrac4{\left(t+c\right)^2}}$$
and
$$y'=\dfrac d{dt}\left[2\cot^{-1}\left(\dfrac2{t+c}-1\right)-t\right]=
\dfrac{\dfrac4{\left(t+c\right)^2}}{1+\left(\dfrac2{t+c}-1\right)^2}-\dfrac{1+\left(\dfrac2{t+c}-1\right)^2}{1+\left(\dfrac2{t+c}-1\right)^2}=$$
$$=\dfrac{\dfrac4{\left(t+c\right)^2}-2+\dfrac4{t+c}-\dfrac4{\left(t+c\right)^2}}{2-\dfrac4{t+c}+\dfrac4{\left(t+c\right)^2}}=
\dfrac{-2+\dfrac4{t+c}}{2-\dfrac4{t+c}+\dfrac4{\left(t+c\right)^2}}$$
We did it!
