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Starting Calculus for Biologists
Simple Trig Antiderivatives and a First Look atSubstitutions
James K. Peterson
Department of Biological Sciences and Department of Mathematical SciencesClemson University
January 31, 2014
Starting Calculus for Biologists
Outline
1 Simple Trigonometric Function Antiderivatives
2 Substitution!
Starting Calculus for Biologists
Abstract
This lecture first goes over the antiderivative idea for the sin andcos functions. Then, we start exploring the idea of using a variablesubstitution in an antiderivative question. We call this simply,using substitution in integration.
Starting Calculus for Biologists
Simple Trigonometric Function Antiderivatives
The simple trigonometric functions sin(t) and cos(t) also havestraightforward antiderivatives as shown in the Theorem below.Let’s just state the results for the big two we are interested in!
Theorem
Antiderivatives of Simple Trigonometric Functions
1 The antiderivative of sin(t) is − cos(t) + C
2 The antiderivative of cos(t) is sin(t) + C
Starting Calculus for Biologists
Simple Trigonometric Function Antiderivatives
Example
Example
Find∫
(2 cos(t) + 5 sin(t)) dt.
Solution∫(2 cos(t) + 5 sin(t)) dt = 2 sin(t) − 5 cos(t) + C .
Starting Calculus for Biologists
Simple Trigonometric Function Antiderivatives
Example
Example
Find∫
(12 cos(t) + 15 sin(t)) dt.
Solution∫(12 cos(t) + 15 sin(t)) dt = 12 sin(t) − 15 cos(t) + C .
Starting Calculus for Biologists
Simple Trigonometric Function Antiderivatives
Example
Example
Find∫
(−4 cos(t) − 5 sin(t)) dt.
Solution∫(−4 cos(t) − 5 sin(t)) dt = −4 sin(t) + 5 cos(t) + C .
Starting Calculus for Biologists
Simple Trigonometric Function Antiderivatives
Example
Example
Find∫
(3 cos(x) + 8 sin(x)) dx .
Solution∫(3 cos(x) + 8 sin(x)) dx = 3 sin(x) − 8 cos(x) + C .
Starting Calculus for Biologists
Simple Trigonometric Function Antiderivatives
Homework 20
20.1 Find∫
(2 cos(y) − 3 sin(y)) dy .
20.2 Find∫
(sin(t) − 5 cos(t)) dt.
20.3 Find∫
(12 cos(t) − 16 sin(t)) dt.
20.4 Find∫
(− cos(x) + sin(x)) dx .
20.5 Find∫
(11 cos(x) + 9 sin(x)) dx .
20.6 Find∫
(8 sin(t)) dt.
20.7 Find∫
(4 cos(t)) dt.
Starting Calculus for Biologists
Substitution!
Now let’s start the processing of figuring out this thing we havecalled substitution. It is really a process where you look atsomething that seems very hard and all of a sudden, there is aclick in your mind and you say “Wow, that is really a disguisedform of a simple thing”. Then once you know that, you are goodto go and you can figure out the solution.Now the thing is how do you learn how to see the simpler thing ina more complex thing? Well, since that is what scientist’s need todo, it certainly is an important idea. And learning to dosubstitutions in antiderivatives starts you on the journey that willhelp you figure that out.So from humble beginnings and a move into more abstractthinking, you get a payoff. You can see the simpler in the complexin places besides a mathy expression. Treasure that skill as it willbe very useful!We will do all this by example!
Starting Calculus for Biologists
Substitution!
Example
Example
Find∫
(t2 + 4) t dt. How would we do this?
Solution
Let u(t) = t2 + 4.
Then dudt = u′(t) = 2t or 1/2 u′(t) = t.
Substitute these into antiderivative we want to find.∫(t2 + 4) t dt =
∫u(t) 1/2 u′(t) dt
Now let’s use the chain rule.
d
dt
(u(t)2
)= 2 u(t) u′(t).
Starting Calculus for Biologists
Substitution!
Example Continued
Solution
This gives
12
ddt
(u(t)2
)= u(t) u′(t).
Now substitute the above into our antiderivative∫(t2 + 4) t dt =
∫1/2 u(t) u′(t) dt =
∫1
2
1
2
d
dt
(u(t)2
)dt
The antiderivative of a derivative is just the function backwith a constant added; so:∫
(t2 + 4) t dt = 1/4
∫d
dt
(u(t)2
)dt
= 1/4 u(t)2 + C = 1/4 (t2 + 4)2 + C
Starting Calculus for Biologists
Substitution!
Example
Example
Let’s do our example, find∫
(t2 + 4) t dt, again using a morestreamlined notation. The first way we did it is very verbose! Thisway is shorter!! How would we do this?
Solution
Let u(t) = t2 + 4.
Then dudt = u′(t) = 2t or 1/2 u′(t) = t.
Write this in differential form 1/2du = t dt.
Starting Calculus for Biologists
Substitution!
Example Continued
Solution
Then substitute into the antiderivative but this time write uinstead of u(t):∫
(t2 + 4) t dt =
∫u 1/2 du = 1/2
∫udu
But the simplest antiderivative of u is 1/2u2. We know ouranswer can have any constant added to it, so we add C at theend. ∫
(t2 + 4) t dt = 1/2
∫u du = 1/4u2 + C
= 1/4(t2 + 4)2 + C
Starting Calculus for Biologists
Substitution!
Example
Example
Find∫
(t2 + 4)2 5t dt. How would we do this?
Solution
Let u(t) = t2 + 4.
Then dudt = u′(t) = 2t or 1/2 u′(t) = t.
Substitute these things into the antiderivative:∫(t2 + 4)2 t 5dt =
∫u(t)2 5/2 u′(t) dt
Now let’s use the chain rule.
d
dt
(u(t)3
)= 3 u(t)2 u′(t).
Starting Calculus for Biologists
Substitution!
Example Continued
Solution
13
ddt
(u(t)3
)= u(t)2 u′(t).
Now substitute these things into the antiderivative∫(t2 + 4)2 5t dt =
∫5/2 u(t)2 u′(t) dt =
∫5
2
1
3
d
dt
(u(t)3
)dt
The antiderivative of a derivative is just the function with aconstant, so:∫
(t2 + 4) t dt = 5/6
∫d
dt
(u(t)3
)dt
= 5/6 u(t)3 + C = 5/6 (t2 + 4)3 + C
Starting Calculus for Biologists
Substitution!
Example
Example
Let’s do our example, find∫
(t2 + 4)2 5t dt, again using a morestreamlined notation.
Solution
Let u(t) = t2 + 4.
Then dudt = u′(t) = 2t or 1/2 u′(t) = t.
Write this in differential form 1/2du = t dt.
Then substitute these things into the antiderivative we wantto find but this time just write u instead of u(t):∫
(t2 + 4)2 5t dt =
∫u2 5/2 du = 5/2
∫u2du
Starting Calculus for Biologists
Substitution!
Example Continued
Solution
But the simplest antiderivative of u2 is 1/3u3. and we add Cat the end.∫
(t2 + 4) t dt = 5/2
∫u2du = 5/6u3 + C
= 5/6(t2 + 4)3 + C
Starting Calculus for Biologists
Substitution!
Example
Example
Let’s find∫
(t2 + 4)4 8t dt using the streamlined notation.
Solution
Let u(t) = t2 + 4. But to save time just write u = t2 + 4.
The differential is du = 2dt so write 1/2du = t dt.
Hence∫(t2 + 4)4 8t dt =
∫u4 8/2 du = 8/2
∫u4du
=8
2
1
5u5 + C = 4/5 (t2 + 4)5 + C .
Starting Calculus for Biologists
Substitution!
Example
Example
Let’s find∫
(t3 + 6)4 8t2 dt using the streamlined notation.
Solution
Let u(t) = t3 + 6. So u = t3 + 6.
The differential is du = 3t2dt so write 1/3du = t2 dt
Hence∫(t3 + 8)4 8t2 dt =
∫u4 8/3 du = 8/3
∫u4du
=8
3
1
5u5 + C =
8
15(t2 + 4)5 + C .
Starting Calculus for Biologists
Substitution!
Example
Example
Let’s find∫
(t4 + 6)2/3 7t3 dt using the streamlined notation.
Solution
u = t4 + 6.
du = 4t3dt so write 1/4du = t3 dt
Hence∫(t4 + 6)2/3 7t3 dt =
∫u2/3 7/4 du = 7/4
∫u2/3du
=7
4
3
5u5/3 + C = 21/20 (t4 + 6)5/3 + C .
Starting Calculus for Biologists
Substitution!
Homework 21
Time to do some yourself using the streamlined notation! Trythese and right down all the steps.
What is u?
What is du?
What is the new integral?
What is the answer in u?
What is the answer in t?
21.1∫
(t2 + 9)4 2t dt
21.2∫
(t2 + 10)4 3t dt
21.3∫
(t3 + 2)4 5t2 dt
21.4∫
(t5 − 3)5 6t4 dt
21.5∫
(t6 + 6)3/4 11 t5 dt