simple trig antiderivatives and a first look at...

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


Outline

1 Simple Trigonometric Function Antiderivatives

2 Substitution!


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.


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



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 .



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 .



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 .



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 .



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.


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!


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).


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


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.


Write this in differential form 1/2du = t dt.


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


Substitution!

Example

Example

Find∫

(t2 + 4)2 5t dt. How would we do this?

Solution

Let u(t) = t2 + 4.


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).


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


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.


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


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


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 .


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 .


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 .


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

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