I apologise for this stupid question -but our school just recently started on integral calculus. We're using Maths Quest. I think my problem regarding this topic is not about what I'm trying to find, but why I am doing so and so -- so far we've just been taught by using 'substitution where the derivative is present in the integral' technique.
So my question is -- would anyone spare me some informational sights or other textbooks, or exaplain, what the intuition is behind this whole process?
Thorium has pretty much explained integration, pretty much the reverse of differentiation and it has a whole lot of applications! Throughout Chapter 5 in Maths Quest you'll learn a whole lot of techniques you can use to integrate functions - things which are beyond the Methods course.
I'll quickly run you through "substitution where the derivative is present in the integral" - I like to look at it as the reverse of the chain rule. When you differentiate a composite function using the chain rule, you differentiate the outside, then multiply it by the derivative of the inside. When you integrate using substitution with this method, you essentially cancel out the derivative, then integrate as normal.
Below is a standard chain rule question:
+5\\let\ u=3x^{2} & \implies & \frac{du}{dx}=6x\\y=\sin(u)+5 & \implies & \frac{dy}{du}=\cos(u)\\\frac{dy}{dx} & = & \frac{dy}{du}\times\frac{du}{dx}=6x\cos(u)=6x\cos(3x^{2})\\\end{eqnarray*})
Now what if we were asked to integrate
)
? We need to simply recognise that the derivative of

is

which is part of the integral and the steps is as follows:
dx\\if\ u=3x^{2} & & \frac{du}{dx}=6x\ \text{(which\ is\ in\ the integral!)}\\\\\frac{du}{dx} & = & 6x\implies dx=\frac{du}{6x}\\sub\ dx=\frac{du}{6x} & and & u=3x^{2}\ into\ integral:\\\int6x\cos(3x^{2})dx & = & \int6x\cos(u)(\frac{du}{6x})=\int cos(u)du=\sin(u)+C\\\therefore \int6x\cos(3x^{2})dx & = & \sin(3x^{2})+C\\\end{eqnarray*})
So there's one of the cool techniques you get to learn as a Specialist student that Methods kids don't get to learn! be proud

hopefully this helps! (although it wasn't really your question haha)