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Example 1 x 2 dx x 2 16 Example 1 x 2 dx x 2 16 We want - - PowerPoint PPT Presentation

Mar 17, 2024 12 likes •565 views

Example 1 x 2 dx x 2 16 Example 1 x 2 dx x 2 16 We want to use the identity: Example 1 x 2 dx x 2 16 We want to use the identity: 1 + tan 2 t = sec 2 t Example 1 x 2 dx x 2 16 We want to

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SLIDE 1
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SLIDE 2
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SLIDE 3

Example 1

  • x2dx

√ x2 − 16

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SLIDE 4

Example 1

  • x2dx

√ x2 − 16 We want to use the identity:

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SLIDE 5

Example 1

  • x2dx

√ x2 − 16 We want to use the identity: 1 + tan2 t = sec2 t

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SLIDE 6

Example 1

  • x2dx

√ x2 − 16 We want to use the identity: 1 + tan2 t = sec2 t We can write the integral as:

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SLIDE 7

Example 1

  • x2dx

√ x2 − 16 We want to use the identity: 1 + tan2 t = sec2 t We can write the integral as:

  • x2dx

4 x

4

2 − 1

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SLIDE 8

Example 1

  • x2dx

√ x2 − 16 We want to use the identity: 1 + tan2 t = sec2 t We can write the integral as:

  • x2dx

4 x

4

2 − 1 So, we make the substitution:

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SLIDE 9

Example 1

  • x2dx

√ x2 − 16 We want to use the identity: 1 + tan2 t = sec2 t We can write the integral as:

  • x2dx

4 x

4

2 − 1 So, we make the substitution: x 4 = sec t

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SLIDE 10

Example 1

  • x2dx

√ x2 − 16 We want to use the identity: 1 + tan2 t = sec2 t We can write the integral as:

  • x2dx

4 x

4

2 − 1 So, we make the substitution: x 4 = sec t ⇒ x = 4 sec t

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SLIDE 11

Example 1

  • x2dx

√ x2 − 16 We want to use the identity: 1 + tan2 t = sec2 t We can write the integral as:

  • x2dx

4 x

4

2 − 1 So, we make the substitution: x 4 = sec t ⇒ x = 4 sec t dx dt = 4 tan t sec t

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SLIDE 12

Example 1

  • x2dx

√ x2 − 16 We want to use the identity: 1 + tan2 t = sec2 t We can write the integral as:

  • x2dx

4 x

4

2 − 1 So, we make the substitution: x 4 = sec t ⇒ x = 4 sec t dx dt = 4 tan t sec t ⇒ dx = 4 tan t sec tdt

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SLIDE 13

Example 1

So, we can make the substitution:

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SLIDE 14

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

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SLIDE 15

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16

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SLIDE 16

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16

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SLIDE 17

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 =

  • x2
  • 16 sec2 t .4 tan t sec t

√ 16 sec2 t − 16

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SLIDE 18

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.

dx

  • 4 tan t sec t

√ 16 sec2 t − 16

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SLIDE 19

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t

  • 16 sec2 t
  • x2

−16

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SLIDE 20

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16

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SLIDE 21

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)
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SLIDE 22

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1

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SLIDE 23

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1 But:

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SLIDE 24

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1 But: 1 + tan2 t = sec2 t

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SLIDE 25

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1 But: 1 + tan2 t = sec2 t ⇒ sec2 t − 1 = tan2 t

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SLIDE 26

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1 But: 1 + tan2 t = sec2 t ⇒ sec2 t − 1 = tan2 t 64 tan t sec3 tdt 4 √ sec2 t − 1

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SLIDE 27

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1 But: 1 + tan2 t = sec2 t ⇒ sec2 t − 1 = tan2 t 64 tan t sec3 tdt 4

✘✘✘✘✘✘ ✿ tan t

√ sec2 t − 1

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SLIDE 28

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1 But: 1 + tan2 t = sec2 t ⇒ sec2 t − 1 = tan2 t 64 tan t sec3 tdt 4

✘✘✘✘✘✘ ✿ tan t

√ sec2 t − 1 = 64 tan t sec3 tdt 4 tan t

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SLIDE 29

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1 But: 1 + tan2 t = sec2 t ⇒ sec2 t − 1 = tan2 t 64 tan t sec3 tdt 4

✘✘✘✘✘✘ ✿ tan t

√ sec2 t − 1 = 64✘✘

✘

tan t sec3 tdt 4✘✘

✘

tan t

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SLIDE 30

Example 1

So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  • x2dx

√ x2 − 16 = 16 sec2 t.4 tan t sec t √ 16 sec2 t − 16 =

  • 64 tan t sec3 tdt
  • 16 (sec2 t − 1)

= 64 tan t sec3 tdt 4 √ sec2 t − 1 But: 1 + tan2 t = sec2 t ⇒ sec2 t − 1 = tan2 t 64 tan t sec3 tdt 4

✘✘✘✘✘✘ ✿ tan t

√ sec2 t − 1 = 64✘✘

✘

tan t sec3 tdt 4✘✘

✘

tan t = 16

  • sec3 tdt
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SLIDE 31

Example 1

So, we ”only” need to solve this trigonometric integral:

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SLIDE 32

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt
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SLIDE 33

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt

We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts):

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SLIDE 34

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt

We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C
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SLIDE 35

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt

We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

Now we only need to substitute back. Remember that our substitution was:

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SLIDE 36

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt

We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4

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SLIDE 37

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt

We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4 We need to have tan t also as a function of x:

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SLIDE 38

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt

We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4 We need to have tan t also as a function of x: tan t =

  • sec2 t − 1
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SLIDE 39

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt

We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4 We need to have tan t also as a function of x: tan t =

  • sec2 t − 1 =
  • x2

16 − 1

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SLIDE 40

Example 1

So, we ”only” need to solve this trigonometric integral: 16

  • sec3 tdt

We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4 We need to have tan t also as a function of x: tan t =

  • sec2 t − 1 =
  • x2

16 − 1 = 1 4

  • x2 − 16
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SLIDE 41

Example 1

So, we need to substitute:

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SLIDE 42

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C
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SLIDE 43

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

sec t = x 4

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SLIDE 44

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

sec t = x 4, tan t = 1 4

  • x2 − 16
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SLIDE 45

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

sec t = x 4, tan t = 1 4

  • x2 − 16

We get that:

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SLIDE 46

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

sec t = x 4, tan t = 1 4

  • x2 − 16

We get that: 16

  • sec3 tdt = 8.x

4.1 4

  • x2 − 16+
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SLIDE 47

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

sec t = x 4, tan t = 1 4

  • x2 − 16

We get that: 16

  • sec3 tdt = 8.x

4.1 4

  • x2 − 16 + 8 ln
  • x

4 + 1 4

  • x2 − 16
  • + C
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SLIDE 48

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

sec t = x 4, tan t = 1 4

  • x2 − 16

We get that: 16

  • sec3 tdt = ✁

✁ ✕

2 8.x

✁

4.1 4

  • x2 − 16 + 8 ln
  • x

4 + 1 4

  • x2 − 16
  • + C
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SLIDE 49

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

sec t = x 4, tan t = 1 4

  • x2 − 16

We get that: 16

  • sec3 tdt = ✁

✁ ✕ ✁ ✁ ✕

1

2 8.x

✁

4.1

✁ ✁ ✕

2

4

  • x2 − 16 + 8 ln
  • x

4 + 1 4

  • x2 − 16
  • + C
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SLIDE 50

Example 1

So, we need to substitute: 16

  • sec3 tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C

sec t = x 4, tan t = 1 4

  • x2 − 16

We get that: 16

  • sec3 tdt = ✁

✁ ✕ ✁ ✁ ✕

1

2 8.x

✁

4.1

✁ ✁ ✕

2

4

  • x2 − 16 + 8 ln
  • x

4 + 1 4

  • x2 − 16
  • + C

16

  • sec3 tdt = x

2

  • x2 − 16 + 8 ln
  • x

4 + 1 4

  • x2 − 16
  • + C
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SLIDE 51

Example 1

So, we can write our final answer as:

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SLIDE 52

Example 1

So, we can write our final answer as:

  • x2dx

√ x2 − 16

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SLIDE 53

Example 1

So, we can write our final answer as:

  • x2dx

√ x2 − 16 = x 2

  • x2 − 16 + 8 ln
  • x

4 + 1 4

  • x2 − 16
  • + C
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SLIDE 54

Example 1

So, we can write our final answer as:

  • x2dx

√ x2 − 16 = x 2

  • x2 − 16 + 8 ln
  • x

4 + 1 4

  • x2 − 16
  • + C
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SLIDE 55