• 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

For anyone that wants a jumpstart, here is the equivalent video on Khan Academy.

We’re going to throw three more integration rules at you for this last section, each dealing with inverse trigonometric functions. Two quick notes: (a) must be a constant greater than (0), and (u) can be any differentiable function.

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

In all three cases, you’re looking for the roots of the two terms in the denominator, and then matching it with the correct rule. Examples 1–3 in the book go over the basic usage, including tricks we’ve seen before, so I won’t repeat them here.

The new trick that appears is completing the square, which allow you to two turn a quadratic into the sum of a squared binomial and a constant. Remember that the term you are adding is (\left(b/2\right)^2).

\[\begin{align} x^2 - 6x + 13 &= x^2 - 6x \colorbox{yellow}{+ 9 - 9} + 13 \\ &= (x-3)^2 + 4 \end{align}\]

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.

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