In this video we solve a long problem using trigonometric substitution. You'll get a feel of trigonometric substitution.

3. Example 1
x2dx
√
x2 − 16

4. Example 1
x2dx
√
x2 − 16
We want to use the identity:

5. Example 1
x2dx
√
x2 − 16
We want to use the identity:
1 + tan2
t = sec2
t

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

13. Example 1
So, we can make the substitution:

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

41. Example 1
So, we need to substitute:

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

51. Example 1
So, we can write our final answer as:

52. Example 1
So, we can write our final answer as:
x2dx
√
x2 − 16

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

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