The pythagorean theorem, as applied to trigonometry says
$\sin^2 \theta + \cos^2 \theta = 1$
This is the key piece of knowledge for these integrals.
The implications are:
$\cos \theta = \pm \sqrt {1-\sin^2 \theta}\\
\sin \theta = \pm \sqrt {1-\cos^2 \theta}\\
\tan^2 \theta + 1 = \sec^2 \theta$
How does this relate to these integrals...
Whenever you see $x^2 + 1$ in someplace inconvenient, like under a radical or in the denominator, you should be thinking of the substitution $x = \tan \theta.$ With this substitution it will become $\tan^2\theta + 1 = \sec^2 \theta$
Similarly, when you see $1-x^2$ you should be thinking $x=\sin\theta$ or $x = \cos \theta$ and the expression becomes $1-\sin^2\theta = \cos^2\theta$
And when you see $x^2 - 1$ it is a bit of a toss up. Sometimes, $x = \sin \theta$ works and sometimes $x = \sec\theta$ works better. It really has to do whether you have reason to think $|x|<1$ (in which case use the sine substitution) or $|x| > 1$ in which case use the secant substitution.
Taking it up a level.
When you see $x^2 + a^2$ then you should be thinking $x = a\tan \theta$ and when you see $a^2 x^2 + b^2$ think $x = \frac {b}{a}\tan \theta$ Finally, when you see $(x+a)^2 + b^2,$ think $x+a = b\tan \theta.$ These will simplify nicely.
Some examples.
The area of a portion of a circle...

The equation of our circle is $x^2 + y^2 = 1$
We want $\int_a^1 \sqrt {1-x^2} \ dx$
start with:
$x = \cos \theta\\
dx = -\sin\theta\ d\theta$
What happens to our limits of integration?
$a = \cos \theta\\
\theta = \arccos a\\
1 = \cos \theta\\
\theta = 0$
$\int_{\arccos a}^{0} \sqrt {1-\cos^2\theta} (-\sin\theta \ d\theta)$
We can reverse the order of integration if we change the sign. $1-\cos^2 \theta = \sin^2\theta$
$\int_0^{\arccos a} \sqrt {\sin^2\theta} (\sin\theta) \ d\theta\\
\int_0^{\arccos a} \sin^2\theta \ d\theta$
Apply a half-angle identity:
$\sin^2\theta = \frac 12 (1-\cos 2\theta)$
$\int_0^{\arccos a} \frac 12 (1-\cos 2\theta) \ d\theta$
$\frac 12 (\theta-\frac 12 \sin 2\theta)|_0^{\arccos a}$
At this point I like to use the double angle identity
$\frac 12 (\theta-\sin \theta\cos \theta)|_0^{\arccos a}$
$\sin \arccos a = \sqrt {1-a^2}$
$\frac 12 (\arccos a - a\sqrt {1-a^2})$
What does this mean geometrically?

The area of the red plus the green is $\frac 12 \theta = \frac 12 \arccos a$
The height of the red triangle is $\sqrt {1-a^2}$ and the area is $\frac 12 a\sqrt {1-a^2}$
One more example
$\int \frac {1}{x^2+x+1} \ dx$
The denominator looks like a bit of a bear. It doesn't factor, if it did, I would suggest partial fractions. As it doesn't we use "completing the square."
$x^2 + x + 1 = (x+\frac 12)^2 + \frac 34$
$\int \frac {1}{(x+\frac 12)^2 + \frac 34} \ dx$
$x+\frac 12 = \sqrt {\frac 34} \tan \theta\\
dx = \sqrt {\frac 34} \sec^2 \theta\ d\theta$
Don't let those radicals scare you, they are just constants.
$\int \frac {\sqrt {\frac 34}\sec^2\theta}{\frac 34 \tan^2\theta + \frac 34} \ d\theta\\
\frac {1}{\sqrt {\frac 34}}\int \frac {\sec^2\theta}{\sec^2\theta} \ d\theta\\
\frac {2}{\sqrt 3} \theta
$
Now we need to reverse the substitution
$x + \frac 12 = \sqrt {\frac 34} \tan \theta\\
\frac {2}{\sqrt 3} (x+\frac 12) = \tan \theta\\
\theta = \arctan (\frac {2\sqrt 3}{3}x + \frac {\sqrt 3}{3})$
$\frac {2\sqrt 3}{3} \arctan (\frac {2\sqrt 3}{3}x + \frac {\sqrt 3}{3})$
I hope this helps.