1. TECHNIQUES OF INTEGRATION8
TECHNIQUES OF INTEGRATION

2. TECHNIQUES OF INTEGRATION
7.3
Trigonometric Substitution
In this section, we will learn about:
The various types of trigonometric substitutions.

3. TRIGONOMETRIC SUBSTITUTION
In finding the area of a circle or an ellipse, an integral of the form arises, where a > 0.
If it were , the substitution would be effective.
However, as it stands, is more difficult.

4. TRIGONOMETRIC SUBSTITUTION
If we change the variable from x to θ by the substitution x = a sin θ, the identity 1 – sin2θ = cos2θ lets us lose the root sign.
This is because:

5. TRIGONOMETRIC SUBSTITUTION
Notice the difference between the substitution u = a2 – x2 and the substitution x = a sin θ.
In the first, the new variable is a function of the old one.
In the second, the old variable is a function of the new one.

6. TRIGONOMETRIC SUBSTITUTION
In general, we can make a substitution of the form x = g(t) by using the Substitution Rule in reverse.
To make our calculations simpler, we assume g has an inverse function, that is, g is one-to-one.

7. INVERSE SUBSTITUTION
Here, if we replace u by x and x by t in the Substitution Rule (Equation 4 in Section 5.5), we obtain:
This kind of substitution is called inverse substitution.

8. INVERSE SUBSTITUTION
We can make the inverse substitution x = a sin θ, provided that it defines a one-to-one function.
This can be accomplished by restricting θ to lie in the interval [-π/2, π/2].

9. TABLE OF TRIGONOMETRIC SUBSTITUTIONS
Here, we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities.

10. TABLE OF TRIGONOMETRIC SUBSTITUTIONS
In each case, the restriction on θ is imposed to ensure that the function that defines the substitution is one-to-one.
These are the same intervals used in Section 1.6 in defining the inverse functions.

11. TRIGONOMETRIC SUBSTITUTION
Example 1
Evaluate
Let x = 3 sin θ, where –π/2 ≤ θ ≤ π/2.
Then, dx = 3 cos θ dθ and
Note that cos θ ≥ 0 because –π/2 ≤ θ ≤ π/2.)

12. TRIGONOMETRIC SUBSTITUTION
Example 1
Thus, the Inverse Substitution Rule gives:

13. TRIGONOMETRIC SUBSTITUTION
Example 1
As this is an indefinite integral, we must return to the original variable x.
This can be done in either of two ways.

14. TRIGONOMETRIC SUBSTITUTION
Example 1
One, we can use trigonometric identities to express cot θ in terms of sin θ = x/3.

15. TRIGONOMETRIC SUBSTITUTION
Example 1
Two, we can draw a diagram, where θ is interpreted as an angle of a right triangle.

16. TRIGONOMETRIC SUBSTITUTION
Example 1
Since sin θ = x/3, we label the opposite side and the hypotenuse as having lengths x and 3.

17. TRIGONOMETRIC SUBSTITUTION
Example 1
Then, the Pythagorean Theorem gives the length of the adjacent side as:

18. TRIGONOMETRIC SUBSTITUTION
Example 1
So, we can simply read the value of cot θ from the figure:
Although θ > 0 here, this expression for cot θ is valid even when θ < 0.

19. TRIGONOMETRIC SUBSTITUTION
Example 1
As sin θ = x/3, we have θ = sin-1(x/3).
Hence,

20. TRIGONOMETRIC SUBSTITUTION
Example 2
Find the area enclosed by the ellipse

21. TRIGONOMETRIC SUBSTITUTION
Example 2
Solving the equation of the ellipse for y, we get or

22. TRIGONOMETRIC SUBSTITUTION
Example 2
As the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant.

23. TRIGONOMETRIC SUBSTITUTION
Example 2
The part of the ellipse in the first quadrant is given by the function
Hence,

24. TRIGONOMETRIC SUBSTITUTION
Example 2
To evaluate this integral, we substitute x = a sin θ.
Then, dx = a cos θ dθ.

25. TRIGONOMETRIC SUBSTITUTION
Example 2
To change the limits of integration, we note that:
When x = 0, sin θ = 0; so θ = 0
When x = a, sin θ = 1; so θ = π/2

26. TRIGONOMETRIC SUBSTITUTION
Example 2
Also, since 0 ≤ θ ≤ π/2,

27. TRIGONOMETRIC SUBSTITUTION
Example 2
Therefore,

28. TRIGONOMETRIC SUBSTITUTION
Example 2
We have shown that the area of an ellipse with semiaxes a and b is πab.
In particular, taking a = b = r, we have proved the famous formula that the area of a circle with radius r is πr2.

29. TRIGONOMETRIC SUBSTITUTION
Note
The integral in Example 2 was a definite integral.
So, we changed the limits of integration, and did not have to convert back to the original variable x.

30. TRIGONOMETRIC SUBSTITUTION
Example 3
Find
Let x = 2 tan θ, –π/2 < θ < π/2.
Then, dx = 2 sec2 θ dθ and

31. TRIGONOMETRIC SUBSTITUTION
Example 3
Thus, we have:

32. TRIGONOMETRIC SUBSTITUTION
Example 3
To evaluate this trigonometric integral, we put everything in terms of sin θ and cos θ:

33. TRIGONOMETRIC SUBSTITUTION
Example 3zz
Therefore, making the substitution u = sin θ, we have:

34. TRIGONOMETRIC SUBSTITUTION
Example 3
We use the figure to determine that:
Hence,

35. TRIGONOMETRIC SUBSTITUTION
Example 4
Find
It would be possible to use the trigonometric substitution x = 2 tan θ (as in Example 3).

36. TRIGONOMETRIC SUBSTITUTION
Example 4
However, the direct substitution u = x is simpler.
This is because, then, du = 2x dx and

37. TRIGONOMETRIC SUBSTITUTION
Note
Example 4 illustrates the fact that, even when trigonometric substitutions are possible, they may not give the easiest solution.
You should look for a simpler method first.

38. TRIGONOMETRIC SUBSTITUTION
Example 5
Evaluate where a > 0.

39. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 1
We let x = a sec θ, where 0 < θ < π/2 or π < θ < π/2.
Then, dx = a sec θ tan θ dθ and

40. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 1
Therefore,

41. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 1
The triangle in the figure gives:

42. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 1
So, we have:

43. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Sol. 1 (For. 1)
Writing C1 = C – ln a, we have:

44. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 2
For x > 0, the hyperbolic substitution x = a cosh t can also be used.
Using the identity cosh2y – sinh2y = 1, we have:

45. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 2
Since dx = a sinh t dt, we obtain:

46. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Sol. 2 (For. 2)
Since cosh t = x/a, we have t = cosh-1(x/a) and

47. TRIGONOMETRIC SUBSTITUTION
E. g. 5—Sol. 2 (For. 2)
Although Formulas 1 and 2 look quite different, they are actually equivalent by Formula 4 in Section 3.11

48. TRIGONOMETRIC SUBSTITUTION
Note
As Example 5 illustrates, hyperbolic substitutions can be used instead of trigonometric substitutions, and sometimes they lead to simpler answers.
However, we usually use trigonometric substitutions, because trigonometric identities are more familiar than hyperbolic identities.

49. TRIGONOMETRIC SUBSTITUTION
Example 6
Find
First, we note that
So, trigonometric substitution is appropriate.

50. TRIGONOMETRIC SUBSTITUTION
Example 6
is not quite one of the expressions in the table of trigonometric substitutions.
However, it becomes one if we make the preliminary substitution u = 2x.

51. TRIGONOMETRIC SUBSTITUTION
Example 6
When we combine this with the tangent substitution, we have
This gives and

52. TRIGONOMETRIC SUBSTITUTION
Example 6
When x = 0, tan θ = 0; so θ = 0.
When x = , tan θ = ; so θ = π/3.

53. TRIGONOMETRIC SUBSTITUTION
Example 6

54. TRIGONOMETRIC SUBSTITUTION
Example 6
Now, we substitute u = cos θ so that du = - sin θ dθ.
When θ = 0, u = 1.
When θ = π/3, u = ½.

55. TRIGONOMETRIC SUBSTITUTION
Example 6
Therefore,

56. TRIGONOMETRIC SUBSTITUTION
Example 7
Evaluate
We can transform the integrand into a function for which trigonometric substitution is appropriate, by first completing the square under the root sign:

57. TRIGONOMETRIC SUBSTITUTION
Example 7
This suggests we make the substitution u = x + 1.
Then, du = dx and x = u – 1.
So,

58. TRIGONOMETRIC SUBSTITUTION
Example 7
We now substitute
This gives and

59. TRIGONOMETRIC SUBSTITUTION
Example 7
So,

60. TRIGONOMETRIC SUBSTITUTION
The figure shows the graphs of the integrand in Example 7 and its indefinite integral (with C = 0).
Which is which?