1. Summary of Trig Substitutions (9/20/13)
Below, a stands for any constant.
“Chunk” is: Substitution: dx is: Chunk is now:
a2 – x2 x = a sin() a cos()d a2 cos2()
a2 + x x = a tan() a sec2()d a2 sec2()
x2 – a x = a sec() a sec()tan()d a2 tan2()
Now replace the chunk, dx, all other occurrences of x with expressions involving , and you are looking at a trig integral in the variable .
If it’s just an antiderivative, go back to x at the end.
If it’s a definite integral, change the endpoints to  .

2. Clicker Question 1 What is ? A. (1/2) (x2 – 9) + C
B. (-1/3)(x2 – 9)-3/2 + C
C. 2(x2 – 9) + C
D. (x2 – 9) + C
E. (1/3)(x2 – 9)-3/2 + C

3. Clicker Question 2 What is ? A. 2(x2 – 9) + C
B. ln(x + (x2 – 9)) + C
C. ln(x – (x2 – 9)) + C
D. ln(x) + ln((x2 – 9)) + C
E. (1/3)(x2 – 9)-3/2 + C

4. Clicker Question 3 What is ? A. 3/2 B.  / 6 C. 10 / (93)
D.  / / (93)
E. 5 / 3

5. Assignment for Monday
Hand-in #1 is due. Be sure to adhere to the Ground Rules.