Regular $u$-substitution works by setting $u=g(x)$ for some function $g$. We have to pick $u$ so that the integrand becomes a nice function of $u$ times $du$. In the video, we explore inverse substitutions.

With inverse substitutions, we let $x = g(\theta)$, where typically $g$ is a trig function. Then $dx = g'(\theta)\, d\theta$, and we can always rewrite the integrand in terms of $\theta$ and $d\theta$. The tricky part is converting everything back to $x$ at the end of the problem.

There are three very useful trig substitution:

For integrals involving $x^2 + a^2$ or $\sqrt{x^2+a^2}$, use $x = a \tan(\theta)$.

Then $dx = a\, \sec^2(\theta)\, d\theta$ and $x^2+a^2 = a^2 \sec^2(\theta)$. To convert back to $x$, draw a right triangle with opposite side $x$, adjacent side $a$ and hypotenuse $\sqrt{x^2+a^2}$ and use soh-cah-toa.

For integrals involving $a^2 - x^2$ or $\sqrt{a^2-x^2}$, use $x=a \sin(\theta)$.

Then $dx = a \cos(\theta) d\theta$ and $\sqrt{a^2-x^2}= a \cos(\theta)$. To convert back to $x$, draw a right triangle with opposite side $x$, hypotenuse $a$ and adjacent side $\sqrt{a^2-x^2}$ and use soh-cah-toa.

For integrals involving $x^2-a^2 $ or $\sqrt{x^2-a^2}$, use $x=a \sec(\theta)$.

Then $dx = a \sec(\theta) \tan(\theta)\, d\theta$ and $\sqrt{x^2-a^2}= a \tan(\theta)$. To convert back to $x$, draw a right triangle with adjacent side $a$, hypotenuse $x$ and opposite side $\sqrt{x^2-a^2}$ and use soh-cah-toa.