Page 30 - Calculus for the Clueless, Calc II

We now have integrals involving square roots. Our goal is to get rid of the radicals. The first area here is trig
substitutions.

Type 1

(a - x ) . We use x = a sin u (dx = a cos u du).
2
2 1/2

and the square root is gone. Here are the other two cases.

Type 2

(a + x ) ; we use x = a tan u (dx = a sec u du)
2
2
2 1/2

Type 3

2 1/2
2
(x - a ) ; we use x = a sec u (dx = a tan u sec u du)

I have demonstrated each of the three types. However, it is essential that you know by sight what the answer is
without substituting. Otherwise, the problems will take forever. If I wake you up in the middle of the night and
ask you, "What do you get if you have (7 - x ) ?" You should instantly say, "Square root of 7 cosine u—now
2 1/2
let me go back to sleep!!!!"

Example 12—

We didn't start with u; we started with x. We must draw a triangle with x = 4 sin u. sin u = x/4. Note the missing
side will be what the square root is.

Example 13—