6. Example 1
x2dx
√
x2 − 16
We want to use the identity:
1 + tan2
t = sec2
t
We can write the integral as:
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
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:
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
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
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
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
14. Example 1
So, we can make the substitution:
x = 4 sec t dx = 4 tan t sec tdt
15. Example 1
So, we can make the substitution:
x = 4 sec t dx = 4 tan t sec tdt
x2dx
√
x2 − 16
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
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
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
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
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
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)
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
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:
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
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
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
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$$$$$$Xtan t√
sec2 t − 1
28. Example 1
So, we can make the substitution:
x = 4 sec t dx = 4 tan t sec tdt
x2dx
√
x2 − 16
=
16 sec2 t.4 tan t sec t
√
16 sec2 t − 16
=
64 tan t sec3 tdt
16 (sec2 t − 1)
=
64 tan t sec3 tdt
4
√
sec2 t − 1
But:
1 + tan2
t = sec2
t ⇒ sec2
t − 1 = tan2
t
64 tan t sec3 tdt
4$$$$$$Xtan t√
sec2 t − 1
=
64 tan t sec3 tdt
4 tan t
29. Example 1
So, we can make the substitution:
x = 4 sec t dx = 4 tan t sec tdt
x2dx
√
x2 − 16
=
16 sec2 t.4 tan t sec t
√
16 sec2 t − 16
=
64 tan t sec3 tdt
16 (sec2 t − 1)
=
64 tan t sec3 tdt
4
√
sec2 t − 1
But:
1 + tan2
t = sec2
t ⇒ sec2
t − 1 = tan2
t
64 tan t sec3 tdt
4$$$$$$Xtan t√
sec2 t − 1
=
64$$$tan t sec3 tdt
4$$$tan t
30. Example 1
So, we can make the substitution:
x = 4 sec t dx = 4 tan t sec tdt
x2dx
√
x2 − 16
=
16 sec2 t.4 tan t sec t
√
16 sec2 t − 16
=
64 tan t sec3 tdt
16 (sec2 t − 1)
=
64 tan t sec3 tdt
4
√
sec2 t − 1
But:
1 + tan2
t = sec2
t ⇒ sec2
t − 1 = tan2
t
64 tan t sec3 tdt
4$$$$$$Xtan t√
sec2 t − 1
=
64$$$tan t sec3 tdt
4$$$tan t
= 16 sec3
tdt
31. Example 1
So, we ”only” need to solve this trigonometric integral:
32. Example 1
So, we ”only” need to solve this trigonometric integral:
16 sec3
tdt
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):
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
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:
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
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:
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
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
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
42. Example 1
So, we need to substitute:
16 sec3
tdt = 8 sec t tan t + 8 ln |sec t + tan t| + C
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
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
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:
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+
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
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
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
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