Differential equation $cos(x-t)=x'$ So we have following equation to solve:
$$\cos(x-t)=x'$$
This is the first time I use substitution, so I want to ensure that I'm doing it correctly, even though it is probably very easy example.
We have:
$$\cos(x-t)=\frac{dx}{dt}$$
We substitute $y=x-t$
$$\cos y=\frac{d(y+t)}{dt}=\frac{dy}{dt}+1$$
So:
$$dt=\frac{dy}{\cos y-1}$$
And now I have to solve integrals:
$$\int dt=\int\frac{dy}{\cos y-1}$$
Is it correct?
 A: I assume that by now, you have solved the remainder of your problem, but for some reason did this post re-emerge for me. So I am just going to pretend this post is not a year old :)
Yes, so far you are correct.
As a next step I'd suggest you make use of the half-angle identity for the sine function, $-2\sin^2\left(\frac{y}{2}\right)=\cos(y)-1:$
$$\int dt=-\int\frac{dy}{2\sin^2\left(\frac{y}{2}\right)}.$$
Then, you can let $u=\cot\left(\frac{y}{2}\right)=\frac{\cos\left(\frac{y}{2}\right)}{\sin\left(\frac{y}{2}\right)}$ and determine $\frac{du}{dy}$ via the quotient rule :
$$\frac{du}{dy}=\frac{d}{dy}\frac{\cos\left(\frac{y}{2}\right)}{\sin\left(\frac{y}{2}\right)}=\frac{\sin\left(\frac{y}{2}\right)\frac{d}{dy}\cos\left(\frac{y}{2}\right)-\cos\left(\frac{y}{2}\right)\frac{d}{dy}\sin\left(\frac{y}{2}\right)}{\sin^2\left(\frac{y}{2}\right)}=-\frac{\sin^2\left(\frac{y}{2}\right)+\cos^2\left(\frac{y}{2}\right)}{2\sin^2\left(\frac{y}{2}\right)}.$$
You can make use of a Pythagorean identity, $1=\sin^2\left(\frac{y}{2}\right)+\cos^2\left(\frac{y}{2}\right):$
$$\frac{du}{dy}=-\frac{1}{2\sin^2\left(\frac{y}{2}\right)}.$$
Substitution into the equation gives
$$\int dt=\int du\rightarrow t+\text{c}_1=u\rightarrow t+\text{c}_1=\cot\left(\frac{y}{2}\right)\rightarrow t+\text{c}_1=\cot\left(\frac{x-t}{2}\right).$$
Rearranging gives an expression for $x$ as a function of $t:$
$$x(t)=2\space\text{arccot}(t+\text{c}_1)+t.$$
