How Trig Substitution Works

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.