Intuition for Integration and Starting Points I've been learning about all the different integration tricks lately, and I'm wondering what your thinking process is when you encounter a new problem.
For example, say we need to simplify$$I=\int\limits_0^a dx\,\frac {\log x}{\sqrt{ax-x^2}}$$The way to go is to notice how replacing $x$ with $a-x$ in the radicand reduces to the same expression. So$$I=\int\limits_0^adx\frac {\log(a-x)}{\sqrt{ax-x^2}}=\int\limits_0^adx\frac {\log(ax-x^2)}{\sqrt{ax-x^2}}$$And adding the two expressions together gives you a nice integral which can be evaluated much easily. However, I'm stuck on how you come up with the idea in the first place. I'm having trouble with finding a good place to start when I encounter these kinds of problems.
Another example is with$$\int\limits_0^1dx\,\log\left(\frac {1+x}{1-x}\right)\frac 1{x\sqrt{1-x^2}}=\frac {\pi^2}2$$If you make a substitution$$t=\frac {1+x}{1-x}\quad\implies\quad x=\frac {1-t}{1+t}$$Then the problem reduces to another problem which can be solved using substitution. However, I'm having trouble understanding how you come up with the substitution for $t$ in the first place.

Question:

*

*What goes through your mind when you're integrating an expression you've never encountered before?

*What are a few useful tricks in integration that may help?


I'm sorry if this post was a bit vague or hard to understand. I'm having a hard time grasping it and putting it into simple, legible English.
 A: $$
\int_0^a \frac{\log x}{\sqrt{ax-x^2}} \, dx = \int_0^a \frac{\log x}{\sqrt{x(a-x)}} \,dx
$$
The number $x$ is the distance from the left endpoint of the interval $[0,a],$ and $a-x$ is the distance from the right enpoint, and that is the interval over which one integrates.
Letting $y = {}$distance from the right endpoint and $a-y={}$distance from the left endpoint, you have the same expression in the denominator since you're just multiplying both distances, and going from one endpoint to the other you have $y$ going from $0$ to $a.$ (You have $dy = -dx,$ and $y$ goes from $a$ down to $0$ as $x$ goes from $0$ up to $a,$ but then the minus sign vanishes when you change $\int_a^0$ to $\int_0^a.$ And you get
$$
\int_0^a \frac{\log(a-y)}{\sqrt{ay-y^2}} \,dy
$$
and that is of course
$$
\int_0^a \frac{\log(a-x)}{\sqrt{ax-x^2}} \, dx.
$$
You misplaced an "equals" sign when you said that is equal to $\displaystyle \int_0^a \frac{\log(ax-x^2)}{\sqrt{ax-x^2}}\,dx.$ Those are not equal; rather the sum of the two integrals is equal to that.
The beta distribution is the probability distribution with density $\text{constant} \times x^{\alpha-1} (1-x)^{\beta-1},$ and its expected value is $\alpha/(\alpha+\beta),$ and in doing various things with that integral, one should remember that $x$ and $1-x$ are both distances from opposite endpoints, so they're both the same kind of thing; there's a symmetry here.
As far as how you think of things like this in the first place, unfortunately with integrals, unlike derivatives, there's no algorithmic rule.
