Can we evaluate $\int_{0}^{t_o} \cos(x(t)) dt $ from $\int_{0}^{t_o} \sin(x(t)) dt$? Can we evaluate $\int_{0}^{t_o} \cos(x(t)) dt $ from $\int_{0}^{t_o} \sin(x(t)) dt$  given that $x$ and $t$ are two variables? Actual dependence of $x$ on $t$ is complicated so that it is supposed to be unknown.
Also it is known that:


*

*$t=0, x(t) = \frac{\pi}{2}$

*$t=t_o, x(t) = 0$


Is it possible to evaluate?
EDIT
The source of this problem is a pursuit curve question, from Problems in General Physics by IE Irodov : archive.org

Point $A$ moves uniformly with velocity $v$ so that the vector $v$ 
  is continually "aimed" at point $B$ which in its turn moves recti- 
  linearly and uniformly with velocity $u < v$. At the initial moment of 
  time $v$ is perpendicular to $u$ and the points are separated by a distance $l$. How soon will the points converge?
  - Problem 13 in Mechanics, Problems in General Physics by IE Irodov

So I supposed that the particles meet at time $t_o$. Also I let the angle made by line joining $A$ and $B$ with the horizontal line as $x(t)$ (angle $x$ is a function of time). SO if the particles are to meet, then:


*

*Distance travelled by $A$ relative to $B$ is simply initial distance between $A$ and $B$, which is $l$.


$$vt_o =l+\int_0^{t_o}u\cos(x(t))dt$$


*Distance travelled by both $A$ and $B$ in horizontal direction must be same if they meet. So using this we get


$$u{t_o}=\int_0^{t_o}v\cos(x(t))dt$$
So this solves our problem.


*Distance travelled by $A$ in horizontal direction in time $t_o$ will simply be initial distance between $A$ and $B$, ie, $l$ :


$$\int_0^{t_o}v\sin(x(t))dt = l$$
We clearly see that Point $1$ and $2$ are sufficient to solve that problem. We take the value of the integral from Point 2 and substitute in Point 1.
My question was that, can we use Point $1$ and $3$ alone to find $t_o$, or what I originally posted:

Can we evaluate $\int_{0}^{t_o} \cos(x(t)) dt $ from $\int_{0}^{t_o} \sin(x(t)) dt$  given that $x$ and $t$ are two variables?

 A: note that $$\sin \left( x \right) =\cos\left(\frac{\pi}{2}-x\right)$$
then we get
$$\int_{0}^t\sin(x)dt=\int_{0}^t\cos\left(\frac{\pi}{2}-x\right)dx=1-\cos(t)$$
A: It would seem highly implausible. A function $x(t)$ that satisfies the properties you mentioned is
$$x(t)=\frac{\pi}{2}\cos(t)$$
with $t_0=\frac{\pi}{2}$. Wolfram Alpha gives us that 
$$\int_0^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\cos(t)\right)dt=\frac{\pi}{2}H_0\left(\frac{\pi}{2}\right)$$
where $H_0$ is the Struve function: https://en.wikipedia.org/wiki/Struve_function, while 
$$\int_0^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\cos(t)\right)dt=\pi J_0\left(\frac{\pi}{2}\right)$$
where $J_0$ is the Bessel function of the first kind: http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html. 
Although you can see that $H_0$ and $J_0$ are both solutions to that differential equation, they are quite different on the right hand side; one is $0$ and one is some gargantuan mess I will not write out. I think that because of this, it is highly implausible that you can do this; although there may be some deep underlying mathematics that I do not understand. Perhaps post your question on the physics site for help with it; I do not know physics so I can't help. ¯_(ツ)_/¯
