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 what is called inverse substitution, which we use for trig substitution. This is
where we substitute something in for $x$, instead of creating a
$u$ that is a function of $x$.

Trig substitution is appropriate when we have an integrand
containing the sum or difference of the squares of a constant and
a variable, i.e. one of the forms $x^2+a^2,
a^2-x^2$, and $x^2-a^2$. Examples integrals for
which we would use trig substitution include those below.
Notice that regular substitution will not work with these
integrals. Also notice it is fine if the $x^2$ has a
coefficient, and in the last example, we must complete the square
in order to get one of our forms. You should begin to
recognize such integrals as trig
substitution integrals.

$$\int\frac{\sqrt{9-x^2}}{x^2}\,dx,\qquad
\int\frac{1}{x^2\sqrt{x^2+4}}\,dx\qquad\int\frac{dx}{(4x^2-25)^{3/2}},\quad\text{
and }\quad \int\frac{x^2}{(3+4x-4x^2)^{3/2}}\,dx$$In trig
substitution, we let $x = g(\theta)$, where $g$ is a trig
function, and then $dx = g'(\theta)\, d\theta$. Since $x$
and $dx$ appear in the integrand, we can always rewrite the
integrand in terms of $\theta$ and $d\theta$. The question
is whether the substitution helps us integrate. Fortunately,
we can teach you how to make good substitutions. It is also
non-trivial to convert everything back to $x$ at the end of the
problem, or in the case of a definite integral, to change your
limits of integration to be in terms of $\theta$.

The above three forms indicate the trig subsitutions we will use,
and they are easy to remember since you know the derivatives of
$\sin^{-1}x,\tan^{-1}x$, and (maybe) $\sec^{-1}x$. These
trig substitutions are derived in the following video, and
summarized on the next page.