• Integrate functions whose antiderivatives involve inverse trigonometric functions.
  • Use the method of completing the square to integrate a function.
  • Review the basic integration rules involving elementary functions.

Assignment

  • Vocabulary and teal boxes
  • p361 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 33–36, 45, 47, 54, 55, 59, 62 70, 74–77

Additional Resources

Integrals and Inverse Trigonometric Functions

We’re going to throw three more integration rules at you for this last section, each dealing with inverse trigonometric functions.

Integrals Involving Inverse Trigonometric Functions

Let $a>0$, and $u$ be any differentiable function.

\[\begin{align} \int \frac {1}{\sqrt{a^2 - u^2}} \, du &= \arcsin{\frac{u}{a}} + C \\ \int \frac {1}{a^2 + u^2} \, du &= \frac{1}{a}\arctan{\frac{u}{a}} + C\\ \int \frac {1}{u\sqrt{u^2 - a^2}} \, du &= \frac{1}{a}\text{arcsec}{\frac{\left| u \right|}{a}} + C \end{align}\]

There are rules for the other three, but they are just the negated versions of the ones above.

Examples 1–3 in the book go over the basic usage, including tricks we’ve seen before, like factoring out a constant or $u$-substitution.

Completing the Square

The new trick that appears in this section is completing the square. In the past, you’ve used this to solve quadratic equations. You would add what you needed to make the quadratic a perfect square, and the same value was added to other side of the equation.

But we are not solving equations, so anything added has to be immediately be subtracted. The result is the sum of a squared binomial and a constant. Remember that the term you are adding is $\left(b/2\right)^2$.

\[\begin{align} & x^2 - 6x + 13 \\ & x^2 - 6x \mathbin{\color{red}{+}}{\color{red}9 - 9} + 13 \\ & (x^2 - 6x + 9) - 9 + 13 \\ & (x-3)^2 + 4 \end{align}\]

All the inverse trig rules involve some combination of $u^2$ and $a^2$, which is what end up with when completing the square. So, here it is in use.

\[\begin{align} \int \frac{1}{x^2 - 4x + 7} \, dx &= \int \frac{1}{x^2 - 4x +4 - 4 + 7} \, dx \\ &= \int \frac{1}{(x - 2)^2 + 3} \, dx \\ &= \frac{1}{\sqrt{3}}\arctan\frac{x-2}{\sqrt{3}} + C \end{align}\]

Here’s another, but now we have sign changes to worry about.

\[\begin{align} \int \frac {1} { \sqrt{3x - x^2} } \, dx &= \int \frac {1} { \sqrt{ -\left ( x^2 - 3x + \frac{9}{4} - \frac{9}{4} \right) } } \, dx \\ &= \int \frac {1} { \sqrt{ -\left(x - \frac{3}{2} \right)^2 + \frac{9}{4} } } \, dx \\ &= \int \frac {1} { \sqrt{ \frac{9}{4} -\left(x - \frac{3}{2} \right)^2 } } \, dx \\ &= \arcsin\frac{x - \frac{3}{2}}{\frac{3}{2}} + C \\ &= \arcsin\left(\frac{2}{3}x - 1 \right) + C \end{align}\]

Review of Basic Integration Rules

The brief, remaining part of the section discusses what you can and can’t integrate with what you have been taught so far. Knowing what you can’t integrate might have limited use on the AP Exam, but the other examples are nice a summary of the forms that you can work with, albeit with a little work to transform them.