I think I see the core of your question from your last paragraph, your image, and your response in post. The essential answer to your question boils down to the definitions of the trig functions.
The definitions of the trigs are nothing more than definitions, and they are defined over right triangles with $\theta$ being either of the non-right angles. That is to say for example, as you surely know, $\sin\theta=\frac{opposite}{hypoteneuse}$. By geometric principles, given any two sides of any right triangle we can always find the third side by the theorem of Pythagoras.
The technique of trigonometric substitution (like any other substitution) works when we can make a total substitution, and then completely reverse substitute the final result. In the case of succesful trig substitutions, we can always go back and call out the value of any of the trigs including $\cot \theta$. We can draw these because ALL of the trigs are defined over right triangles, and we can always find a third side given two. No one trig is more difficult to find than another over a right triangle with defined side lengths.
When all of the terms of an integral can be completely substituted by the method, the result, assuming you can find it, should always be re-translatable in to the original terms as all of the information required for the back substitution is included in the contents of a right triangle, as a trig of an angle is nothing more than the ratio of two sides of a right triangle.
Given any of the 6 trigs with variable theta, we can always assign measures to two of three sides of a right triangle as that is how the trigs are defined, hence we can always define the third side.
I hope this post is helpful, please let me know if it is less than clear. Trig sub is my favorite technique as far as methods go and so I am happy to have this out with you.
You always can use the substitution $t=\tan \dfrac x2$ to compute the integral of a rational function of trigonometric functions.
Now Bioche's rules tell you can use the simpler substitution $ u=\sin x$.
Best Answer
The point is that for definite integrals you just need your substitution to be "piecewise injective", since you can split your integrals in finitely many parts on which your piecewise injective substitution is injective. For instance, the substitution $x^2 = t$ is injective if you restrict $x$ to $[0,\infty[$ or $]-\infty,0]$, so essentially you can use this change of variables everywhere. Same goes with any polynomial substitution $t = p(x)$ ; the polynomial $p'(x)$ has finitely many zeros, hence the derivative of $p(x)$ can change sign only finitely many times, thus is piecewise injective.