Handout Research shows that students learn when they MAKE MISTAKES and not when they get it right. SO MAKE MISTAKES, because MISTAKES ARE GOOD! Review:. The Substitution Rule Indefinite f (g()) g ()d = f(u)du Definite b a f (g()) g ()d = g(b) g(a) f(u)du Parameters: f must be continuous on the range of u = g() and either g () is continous on the interval [a,b] (Definite Case) or g() is differentiable (Indefinite Case).. Useful Integrals sec ()d = tan() + C sec()tan()d = sec() + C cosh()d = sinh() + C sinh()d = cosh() + C ( a d = ) sin + C a + a d = ( ) a tan + C a cot()d = ln sin() + C tan()d = ln sec() + C a d = a ln(a) + C d = ln + C 3. Even and Odd Functions A function f is even if f( ) = f() for every in the domain of f. Eamples:, 4, cos() over R. A function f is odd if f( ) = f() for every in the domain of f. Eamples:, 3, sin() over R. 4. Eamples: Notes: The purpose of the Substitution Rule is to convert an unfamiliar integral into one that is familiar. The trick to using the rule successfully is to look for a function, possibly composed within another function, with its derivative laying around. For instance, in the integral ( in the book) sin(ln()) d, we see the natural log sitting inside the sine function, and its derivative laying below it. So we might try using Substitution as follows: Let u = ln() then du = d and we have sin(ln()) d = = [ ] sin ([ ln()]) d sin([u])[du] = cos([u]) + C = cos([ln()]) + C
Eample II: Number 36 in book. Hint: Which of the above integrals does it look like? + 4 d Eample III: Number 48 in the book. 4 0 + d
5. Integration by Parts Reversing the product rule Let f and g be differentiable functions, then the product rule states d d [f()g()] = Integrating both sides and splitting the integral implies: f()g() = If we let u = f(), and v = g() then du = f ()d, and dv = g ()d. With a little rearranging we get udv = uv vdu 6. Eamples: Notes: The trick to using Int. by Parts efficiently is a wise choice of u and dv. That wise choice comes down to intuitively choosing your u as the function that will simplify and eventually go away after repeated differentiation (like k for some natural number k), and a dv that wont get too messy when you integrate it (you have to know how to integrate it of course). Sometimes we use integration by parts on single functions that we don t know the integral of, but we know their derivatives as the following eample shows. Let u = sin () and dv = d, then du = and v =. Thus, 0 sin ()d = [ sin () ] 0 d Using U-sub with u = and du = we get, [ π 0 ] () (0) π ( ) d = π 0 du u [ ] u = π 0 [ 0] = π 0 Eample V: Number 5 in the book. Hint: Try using U-Sub first. e ( + ) d 3
Eample VI: Hint: You might have to use Int. by Parts more than once and look for patterns. e sinh()d 4
Name: In Class Eercises: Write only your solutions on this paper. No Scratch Work! Work in groups of no more than 3.. Independently (Everyone in the group has their own problems) reverse engineer (create) two distinct integrals (not constant multiples of each other) that can be solved using a substitution involving tan( ). Then show their solutions. 5
. As a group (everyone creates the same problem) reverse engineer an integral that can be solved using two different substitutions and show its solution. 6
3. Independently solve the following integrals. Hint: After using Integration by Parts use the Fundamental Trig Id. sin () + cos () =. a.) cos ()d Hint: Use the Fundamental Trig Id., then U-Sub (Int. by parts is not strictly necessary). b.) cos 3 ()d 7
4.) As a group develop a rule for calculating the integral cos n ()d when n N and bigger than. Hint: Try it for n = 4 and n = 5, then look for patterns from 3a and 3b. You might even choose to break it up into two rules, one for when n is even and the other when n is odd, though this is not necessary. 8