This document discusses techniques for trigonometric substitution when integrating expressions involving trigonometric functions. It presents three main strategies for trigonometric substitution: 1) If √a2 - x2 appears, substitute x = a sin θ; 2) If √x2 + a2 appears, substitute x = a tan θ; 3) If √x2 - a2 appears, substitute x = a sec θ. Examples are provided to illustrate each strategy and show how the trigonometric identities can be used to simplify the integrand. A table summarizes the substitutions, their associated θ ranges, and identities.

1. Chapter 7: Techniques of Integration
Section 7.3: Trigonometric Substitution
Alea Wittig
SUNY Albany

2. Outline
Area of Circle of Radius r
Inverse Substitution
Trigonometric Substitution
Strategy 1: x = a sin θ
Strategy 2: x = a tan θ
Strategy 3: x = a sec θ
Table of Trigonometric Substitutions

3. Area of Circle of Radius r

4. ▶ A circle of radius r centered at the origin has the equation
x2
+ y2
= r2
▶ To find the area of the circle, we can compute the area under
the curve y =
√
r2 − x2 from x = 0 to x = r, and then
multiply by 4.
Area of circle = 4
Z r
0
p
r2 − x2dx

5. 4
Z r
0
p
r2 − x2dx
▶ This integral is difficult to do using a normal u substitution.
▶ We know that
sin2
θ + cos2
θ = 1 =⇒ 1 − sin2
θ = cos2
θ
for all θ, so what if we let x = r sin θ?
p
r2 − x2 =
q
r2 − (r sin θ)2
=
p
r2 − r2 sin2
θ
=
q
r2(1 − sin2
θ)
=
√
r2 cos2 θ
= r| cos θ|

6. ▶ If we restrict θ so that −π
2 ≤ θ ≤ π
2 , that corresponds to x
ranging from −r to r and cos θ ≥ 0 for all values of θ in the
interval. So | cos θ| = cos θ for −π
2 ≤ θ ≤ π
2 .
▶ Now
√
r2 − x2 = r cos θ where we are assuming −π
2 ≤ θ ≤ π
2
▶ So this substitution helped us to eliminate the square root in
the integrand which was causing us problems, but we aren’t
done yet.
▶ We also have to determine dx and the limits of integration:
x = r sin θ =⇒
dx = r cos θdθ
x1 = 0 =⇒ r sin θ1 = 0 =⇒ sin θ1 = 0 =⇒ θ1 = sin−1
0 = 0
x2 = r =⇒ r sin θ2 = r =⇒ sin θ2 = 1 =⇒ θ2 = sin−1
1 = π
2
4
Z 1
0
p
r2 − x2dx = 4
Z π
2
0
r cos θr cos θdθ
= 4
Z π
2
0
r2
cos θdθ

7. ▶ Using the half angle formula cos2 θ = 1
2 1 + cos 2θ
we have
4
Z π/2
0
r2
cos2
θdθ = 4r2
Z π/2
0
1
2
(1 + cos 2θ)dθ
= 2r2
Z π/2
0
(1 + cos 2θ)dθ
▶ Let u = 2θ, so du = 2dθ and du
2 = dθ.
▶ θ1 = 0 =⇒ u1 = 0
▶ θ2 = π
2 =⇒ u2 = π
2r2
Z π/2
0
(1 + cos 2θ)dθ =
2r2
Z π
0
(1 + cos u)
du
2
= r2
Z π
0
(1 + cos u)du

8. r2
Z π
0
(1 + cos u)du = r2
u + sin u

11. π
0
= r2
π + sin π − 0 + sin 0
= πr2
▶ So this checks out since we already know the area of a circle of
radius r is πr2.
▶ But we haven’t actually checked that the substitution
x = r sin θ is even well defined!

12. Inverse Substitution

13. Inverse Substitution
▶ Pay careful attention to the difference between the two
substitutions we made:
▶ The substitution x = r sin θ.
▶ We defined the original variable x as a function of a new
variable θ.
▶ This is called inverse substitution.
▶ The substitution u = 2θ.
▶ We defined a new variable u as a function of the original
variable θ.
▶ This is just regular u substitution.

14. Inverse Substitution
▶ Consider the integral Z
f (x)dx
▶ If we make the substitution x = g(t), then dx = g′(t)dt and
the integral becomes
Z
f (g(t))g′
(t)dt
▶ As long as g(t) has an inverse function g−1(t), this
substitution is equivalent to the substitution t = g−1(x) since
by definition of an inverse function,
x = g(t) ⇐⇒ g−1
(x) = t
▶ So as long as the function g(t) has an inverse, ie, is one-to-one
on its domain, then inverse substitution is well defined.

15. Inverse Substitution
▶ Recall that we restrict the domain of sin θ to −π
2 ≤ θ ≤ π
2 in
order to define its inverse function.
▶ We do this as a convention since sin θ is one-to-one and takes
all the values in its range on the interval [−π
2 , π
2 ].
▶ So as long as we are restricting θ so it is contained in the
interval [−π
2 , π
2 ], as we did in the preceding example, the
substitution x = r sin θ is well defined.

16. Trigonometric Substitution

17. Trigonometric Substitution
▶ In this course we will be doing are certain type of inverse
substitution, Trigonometric Substitutions.
▶ There are three principle kinds of trig subs.
▶ We will introduce each by means of a typical example.

18. Strategy 1: x = a sin θ

19. Example 1
1/5
Z √
9 − x2
x2
dx
▶ We will use the same technique as in our solution to the area
of a circle problem.
▶ Let x = 3 sin θ where −π
2 ≤ θ ≤ π
2
▶ that is, θ = sin−1 x
3
▶ Then dx = 3 cos θdθ and
p
9 − x2 =
q
9 − (3 sin θ)2
=
p
9 − 9 sin2
θ
=
q
9(1 − sin2
θ)
= 3
p
1 − sin2
θ
= 3
√
cos2 θ
= 3| cos θ|

20. Example 1
3/5
▶ cos θ ≥ 0 for −π
2 ≤ θ ≤ π
2 so | cos θ| = cos θ.
▶ So we have x = 3 sin θ =⇒
p
9 − x2 = 3 cos θ
x2
= 9 sin2
θ
dx = 3 cos θdθ =⇒
Z √
9 − x2
x2
dx =
Z
3 cos θ
9 sin2
θ
3 cos θdθ
=
Z
cos2 θ
sin2
θ
dθ
=
Z
cot2
θdθ
=
Z
(csc2
θ − 1)dθ
= − cot θ − θ + C

21. Example 1
4/5
▶ Now we have to return to the variable x.
x = 3 sin θ =⇒ sin θ =
x
3
=
opposite
hypoteneuse
so we have a right angle with central angle θ,
▶ opposite side = x
▶ hypoteneuse = 3
▶ adjacent side is then
√
9 − x2 by Pythagorean theorem.
θ
3
√
9 − x2
x

22. Example 1
5/5
▶ So cot θ = adjacent
opposite =
√
9−x2
x and θ = sin−1 x
3
.
▶ We could’ve also use the fact that cot θ = 3 cos θ
3 sin θ =
√
9−x2
x
from earlier calculations.
▶ Now we have
Z √
9 − x2
x2
dx = −
√
9 − x2
x
− sin−1
x
3
+ C

23. Strategy 1
Example 1 illustrates the following general method:
If
p
a2 − x2 appears in the integrand, try the substitution x = a sin θ

24. Strategy 2: x = a tan θ

25. Example 2
1/6
Z
dx
x2
√
x2 + 4
▶ The expression is of the form
√
x2 + a2.
▶ We know that
tan2
θ + 1 = sec2
θ
for all θ, so what if we let x = 2 tan θ?
▶ tan θ is one-to-one and takes all of the values in its range on
the interval (−π
2 , π
2 ) so we restrict θ so it is contained in
(−π
2 , π
2 )

26. Example 2
2/6
p
x2 + 4 =
q
(2 tan θ)2 + 4
=
p
4 tan2 θ + 4
=
q
4(tan2 θ + 1)
=
√
4 sec2 θ
= 2| sec θ|
▶ sec θ ≥ 0 on the interval (−π
2 , π
2 ) so | sec θ| = sec θ.

27. Example 2
3/6
▶ So we have x = 2 tan θ =⇒
p
x2 + 4 = 2 sec θ
x2
= 4 tan2
θ
dx = 2 sec2
θdθ =⇒
Z
dx
x2
√
x2 + 4
=
Z
2 sec2 θdθ
4 tan2 θ · 2 sec θ
=
Z
sec θ
tan2 θ
dθ
=
Z
1
cos θ
cos2 θ
sin2
θ
dθ
=
Z
cos θ
sin2
θ
dθ

28. Example 2
4/6
▶ To compute
R cos θ
sin2
θ
dθ, we could do a u sub with u = sin θ or
we could use trig identities to write
Z
cos θ
sin2
θ
dθ =
Z
cos θ
sin θ
1
sin θ
dθ
=
Z
cot θ csc θdθ
= − csc θ + C
▶ Now to return to the variable x where x = 2 tan θ, we need to
figure out what csc θ is in terms of x.

29. Example 2
5/6
x = 2 tan θ =⇒ tan θ =
x
2
=
opposite
adjacent
so we have a right angle with central angle θ,
▶ opposite side = x
▶ adjacent side = 2
▶ hypoteneuse is then
√
x2 + 4 by Pythagorean theorem.
θ
√
x2 + 4
2
x

30. Example 2
6/6
▶ So csc θ = hypoteneuse
opposite =
√
x2+4
x
▶ We could’ve also used the fact that csc θ = sec θ
tan θ ,
tan θ = x
2 , and sec θ =
√
x2 + 4 from earlier calculations.
▶ Now we have
Z
dx
x2
√
x2 + 4
= −
√
x2 + 4
x
+ C

31. Strategy 2
Example 2 illustrates the following general method:
If
p
x2 + a2 appears in the integrand, try the substitution x = a tan θ

32. Strategy 3: x = a sec θ

33. Example 3
1/4
Z
x2dx
√
x2 − 4
▶ The expression is of the form
√
x2 − 4.
▶ We know that
tan2
θ + 1 = sec2
θ =⇒ sec2
θ − 1 = tan2
θ
for all θ, so what if we let x = 2 sec θ?
▶ sec θ is one-to-one and takes all of the values in its range on
the set [0, π
2 ) ∪ [π, 3π
2 ).

34. Example 3
2/4
p
x2 − 4 =
q
(2 sec θ)2 − 4
=
p
4 sec2 θ − 4
=
q
4(sec2 θ − 1)
= 2
p
sec2 θ − 1
= 2
√
tan2 θ
= 2| tan θ|
▶ tan θ ≥ 0 on the intervals [0, π
2 ) and [π, 3π
2 ) so | tan θ| = tan θ

35. Example 3
3/4
▶ So we have x = 2 sec θ =⇒
p
x2 − 4 = 2 tan θ
x2
= 4 sec2
θ
dx = 2 sec θ tan θdθ =⇒
Z
x2dx
√
x2 − 4
=
Z
4 sec2 θ2 sec θ tan θdθ
2 tan θ
= 4
Z
sec3
θdθ
= 4
1
2
sec θ tan θ + ln | sec θ + tan θ| + C
= 2(sec θ tan θ + ln | sec θ + tan θ| + C1)
from example 10 of 7.2 lecture slides (or example 8 of 7.2 in
the textbook).

36. Example 3
4/4
▶ Now to return to the variable x where x = 2 sec θ we must
figure out what tan θ is in terms of x.
▶ You could form a right triangle, but we already have:
2 tan θ =
p
x2 − 4 =⇒ tan θ =
√
x2 − 4
2
x = 2 sec θ =⇒ sec θ =
x
2
Z
x2dx
√
x2 − 4
= 2(sec θ tan θ + ln | sec θ + tan θ| + C1)
= 2
x
√
x2 − 4
4
+ ln

39. x +
√
x2 − 4
2

42. + C1
!
=
x
√
x2 − 4
2
+ 2 ln |x +
p
x2 − 4| + C
where C = C1 − 2 ln 2 (you can also just wait to add constant of
integration until you are done with the integral)

43. Strategy 3
Example 3 illustrates the following general method:
If
p
x2 − a2 appears in the integrand, try the substitution x = a sec θ

44. Table of Trigonometric Substitutions

45. Table of Trigonometric Substitutions
▶ If you have an integral with one of the following expressions
and a more rudimentary approach (eg, u substitution) won’t
simplify the integral then
▶ make the associated trigonometric substitution
▶ and then use the associated identity to simplify the expression;
the square root should cancel out.
Expression Substitution θ Range Identity
√
a2 − x2 x = a sin θ [−π
2 , π
2 ] 1 − sin2
θ = cos2 θ
√
x2 + a2 x = a tan θ (−π
2 , π
2 ) tan2 θ + 1 = sec2 θ
√
x2 − a2 x = a sec θ [0, π
2 ] ∪ [π, 3π
2 ) sec2 θ − 1 = tan2 θ

46. Remarks
▶ Notice that
▶ cos θ ≥ 0 on [−π
2 , π
2 ]
▶ sec θ ≥ 0 on (−π
2 , π
2 )
▶ tan θ ≥ 0 on [0, π
2 ] ∪ [π, 3π
2 )
▶ So, on their respective intervals, each substitution simplifies
the respective expression, and has differential according to the
following table
Substitution Simplified Expression Differential
x = a sin θ
√
a2 − x2 = a cos θ dx = a cos θdθ
x = a tan θ
√
x2 + a2 = a sec θ dx = a sec2 θdθ
x = a sec θ
√
x2 − a2 = a tan θ dx = a sec θ tan θdθ