How would I "see" this or go a head to solve it?
I just can't understand it or see why I would go that way; how can I make myself able to see stuff like this?
$$\int \frac{t}{ \sqrt{4-t^4}}dt $$
Answer will be that I substitute $t^2 = u$ and then work from there on out, but how would I ever see it that specific?
On
The idea is to reverse the chain rule. If you have $$\int \frac{t}{ \sqrt{4-t^4}}dt,$$ for example, you're motivated to take $u = t^2$ because of the $t$ in the numerator. Then $du = 2t \ dt$, and you almost have it! Making a substitution when the derivative of the object is sitting near is usually works, for example:
etc.
On
If you make a substitution $u = f(t)$, then $du = f'(t) \,dt$, so we can view a choice of substitution as a choice of which factor of the integrand to pair with $dt$, that is, what to take as $f'(t)$; then we can recover what $u$ is by integrating. Essentially your only choices are $\frac{1}{\sqrt{4 - t^4}}$, which looks no easier to integrate than the integrand at hand (and in fact is much harder) and $t$. By elimination, we take $$du = 2 t \,dt$$ so that $$u = t^2.$$ (Here, we've harmlessly introduced a factor of $2$ to prevent the unnecessary appearance of fractional expressions after the substitution---if this isn't apparent, try substituting $u = \frac{1}{2} t^2$, $du = t \,dt$ and see what happens.)
Like you say, with this substitution in hand, the integral becomes $$\int \frac{t \, dt}{\sqrt{4 - t^4}} = \frac{1}{2} \int \frac{du}{\sqrt{4 - u^2}},$$ which can be handled with a standard trigonometric substitution.
With practice, you'll develop the intuition you seek. Picking $u = t^2$ is very helpful here, because $$u = t^2 \implies du = 2t\,dt \iff t\,dt = \dfrac {du}{2}$$
Note that we have $\frac 12du = t\,dt $ in the numerator!
Thus, the integral, after substitution, becomes $$\frac 12\int \frac{ du}{\sqrt {4-u^2}}$$
Now, it's just a matter of using trig substitution $u =2 \sin \theta$.