Theory

Theory 1

Certain algebraic expressions have a secret meaning that comes from the Pythagorean Theorem. This meaning has a very simple expression in terms of trig functions of a certain angle.

For example, consider the integral:

$$\int \frac{1}{x^2\sqrt{x^2-9}}dx$$

Now consider this triangle:

The triangle determines the relation $x=3\sec \theta$, and it implies $\sqrt{x^2-9}=3\tan \theta$.

Now plug these into the integrand above:

$$\frac{1}{x^2\sqrt{x^2-9^2}}\gg \gg \frac{1}{9{\sec }^2\theta \cdot 3\tan \theta }$$

Considering that $dx=3\sec \theta \tan \theta d\theta$, we obtain a very reasonable trig integral:

$$\begin{array}{cc}\int \frac{1}{x^2\sqrt{x^2-9^2}}dx\gg \gg & \int \frac{3\sec \theta \tan \theta }{27{\sec }^2\theta \tan \theta }d\theta \\ & \\ \gg \gg & \frac{1}{9}\int \cos \theta d\theta \gg \gg \frac{1}{9}\sin \theta +C\end{array}$$

We must rewrite this in terms of $x$ using $x=3\sec \theta$ to finish the problem. We need to find $\sin \theta$ assuming that $\sec \theta =\frac{x}{3}$. To do this, refer back to the triangle to see that $\sin \theta =\frac{\sqrt{x^2-9}}{x}$. Plug this in for our final value of the integral:

$$\frac{1}{9}\sin \theta +C\gg \gg \frac{\sqrt{x^2-9}}{9x}+C$$

Here is the moral of the story:

Pythagorean expressions

Re-express the Pythagorean expression using a triangle and a trig substitution.

In this way, we are able to eliminate square roots of quadratics.

There are always three steps for these trig sub problems:

• (1) Identify the trig sub: find the sides of a triangle and relevant angle $\theta$.
• (2) Solve a trig integral (often a power product).
• (3) Refer back to the triangle to convert the answer back to $x$.

To speed up your solution process for these problems, memorize these three transformations:

(1)

$$\sqrt{a^2-x^2}{\gg \gg }^{x=a\sin \theta }\sqrt{a^2-a^2{\sin }^2\theta }=a\cos \theta \text{from}1-{\sin }^2\theta ={\cos }^2\theta$$

(2)

$$\sqrt{a^2+x^2}{\gg \gg }^{x=a\tan \theta }\sqrt{a^2+a^2{\tan }^2\theta }=a\sec \theta \text{from}1+{\tan }^2\theta ={\sec }^2\theta$$

(3)

$$\sqrt{x^2-a^2}{\gg \gg }^{x=a\sec \theta }\sqrt{a^2{\sec }^2\theta -a^2}=a\tan \theta \text{from}{\sec }^2\theta -1={\tan }^2\theta$$

For a more complex quadratic with linear and constant terms, you will need to first complete the square for the quadratic and then do the trig substitution.