Gavin's Calc II Class Clarification: Nov 19

The first question is easy: we need to know one trig identity, and a couple of algebraic (that means we add, subtract, and divide things) extensions of them. The identity is sin²(x) + cos²(x) = 1
and the things we'll do to it are: (1) subtract one of the sine or cosine over to the other side of the equation, (2) divide it by cos²(x), and (3) take the result of (2) and subtract a term. These give sin²(x) = 1 - cos²(x) or cos²(x) = 1 - sin²(x),
tan²(x) + 1 = 1/cos²(x), and
tan²(x) = -1 + 1/cos²(x).

This leads to the second question: how can I make using them more intuitive?

First, notice that the three identities we just derived have the forms

(something)² = 1 - (somethingelse)²,
(something)² = (somethingelse)² + 1, and
(something)² = (somethingelse)² - 1. Replace the "somethingelse" with "x", and we have (something)² = 1 - x²,
(something)² = x² + 1, and
(something)² = x² - 1. So whenever you see a term like one of those on the right-hand side of one of these three, you should think "maybe trig substitution will work here!" Then, to make it work, we have to go from these right-hand sides back to the three identities in the previous paragraph. Therefore, we need to have in mind the following set of substitutions

term in integral	substitution	to get	which is the same as
1 - x2	x = sin(z)	1 - sin2(z)	cos2(z)
1 + x2	x = tan(z)	1 + tan2(z)	1/cos2(z)
x2 - 1	x = 1/cos(z)	-1 + 1/cos2(z)	tan2(z)

After that, the substitution and integration is the same as we have done in the past: find what dx is and use that to replace the dx in the integral with a dz, and then integrate.

Once you've found the integral, convert the antiderivative back to be in terms of x by using the substitution that you originally made. You know that if

x = sin(z), then
arcsin(x) = z. To find other trigonometric forms in terms of x, you can remember the conversions shown in the table below, or, better, remember that cos(z) = (adjacent) / (hypotenuse), sin(z) = (opposite) / (hypotenuse), and tan(z) = (opposite) / (adjacent), and use this to figure out what the other trig functions are. As an example, if we said x = tan(z), then we can say that the side opposite the angle z is x, and that adjacent is just 1 (see figure to right). Thus the hypotenuse must be sqrt(x² + 1), and we can find the remaining trig functions by using the definitions we just mentioned above. This method is used to generate the following table:

substitution	trig function	expression in x
x = sin(z)	sin(z)	x
	cos(z)	sqrt(1 - x2)
	tan(z)	x / sqrt(1 - x2)
x = tan(z)	sin(z)	x / sqrt(1 + x2)
	cos(z)	1 / sqrt(1 + x2)
	tan(z)	x
x = 1/cos(z)	sin(z)	sqrt(x2 - 1) / x
	cos(z)	1 / x
	tan(z)	sqrt(x2 - 1)