Download Trig Substitutions

Trig Substitutions
Trig Substitutions are used to evaluate integrals involving the expressions
a2 − x2 , a2 + x2 or x2 − a2 . Trig
R
Substitutions are a type of ‘reverse substitution’. To evaluate f (x) dx via a reverse substitution we let
x = g(t) so that dx = g 0 (t)dt and
Z
Z
f (g(t))g 0 (t) dt.
f (x) dx =
Trigonometric Substitutions
Substitution
Expression
Identity
a2 − x2
x = a sin(θ) with − π2 ≤ θ ≤
π
2
1 − sin2 (θ) = cos2 (θ)
a2 + x2
x = a tan(θ) with − π2 ≤ θ ≤
π
2
1 + tan2 (θ) = sec2 (θ)
x2 − a2
x = a sec(θ) with 0 ≤ θ <
π
2
or π ≤ θ <
3π
2
sec2 (θ) − 1 = tan2 (θ)
The table indicates the appropriate Trig Substitution effective for the given expression along with a corresponding identity to simplify the integrand. The range of θ-values guarantees the function used for the
substitution is one-to-one.
Z
1
dx.
Example 1. Evaluate
3/2
(x2 + 1)
1
Intuition: The integrand
is not easily evaluated with IBPs or a u-Substitution. However, the
3/2
2
(x + 1)
expression x2 + 1 appearing in the denominator suggests we try the Trig Substitution x = tan(θ).
Let x = tan(θ) so that dx = sec2 (θ)dθ.
Z
Z
1
(x2
+ 1)
3/2
dx =
1
tan2 (θ) + 1
sec2 (θ)
Z
=
3/2
(sec2 (θ))
Z
sec2 (θ)
=
dθ
sec3 (θ)
Z
= cos(θ) dθ
2
3/2 sec (θ) dθ
dθ
= sin(θ) + C
x
+ C.
=√
x2 + 1
Observations:
• Immediately after applying the Trig Substitution use the corresponding identity to simplify the integrand.
• The Trig Substitution converts the original integral into a trigonometric integral.
1
Calculus II Resources
Integration Techniques
• To rewrite the solution in terms of the variable x, draw the right triangle corresponding to the Trig
Substitution (see above).
See solution video
Z √
Example 2. Evaluate
x2 − 4
dx.
x
Intuition: The expression x2 − 4 indicates that we can apply the Trig Substitution x = a sec(θ) with a = 2.
Let x = 2 sec(θ) so that dx = 2 sec(θ) tan(θ)dθ.
Z √
x2 − 4
dx =
x
=
=
Z p
Z p
Z q
Z
=2
4 sec2 (θ) − 4
2 sec(θ) tan(θ) dθ
2 sec(θ)
4(sec2 (θ) − 1) tan(θ) dθ
4 tan2 (θ) tan(θ) dθ
tan2 (θ) dθ
To evaluateR this trigonometric integral we use the technique described in part (3) of the Guidelines for
evaluating tanm (x) secn (x) dx.
Z √
x2 − 4
dx = 2
x
Z
Z
=2
tan2 (θ) dθ
sec2 (θ) − 1 dθ
= 2 tan(θ) − 2θ + C
x
p
= x2 − 4 − 2 sec−1
+ C.
2
Observation: Evaluating integrals requires patience and perseverance. You may have to try multiple approaches.
See solution video
Z
Example 3. Evaluate
2
√
2
x3
√
1
dx.
x2 − 1
Intuition: When applying a Trig Substitution to a definite integral you MUST change the limits of integration.
√
x
√= 2
Let x = sec(θ) so that dx = sec(θ) tan(θ)dθ and
2 = sec(θ)
θ = π4
2
x=2
2 = sec(θ)
θ = π3
Calculus II Resources
Integration Techniques
Z
2
√
2
x3
√
1
dx =
x2 − 1
Z
π/3
π/4
Z
π/3
=
π/4
Z
1
p
sec(θ) tan(θ) dθ
sec3 (θ) sec2 (θ) − 1
1
p
tan(θ) dθ
sec2 (θ) tan2 (θ)
π/3
cos2 (θ) dθ
=
π/4
1
=
2
Z
π/3
1 + cos(2θ) dθ
π/4
π/3
1
1
=
θ + sin(2θ) 2
2
π/4
π 1
π 1
2π
π 1
=
+ sin
−
+ sin
2
3
2
3
4
2
2
"
#
√
1 π
1 3 1
=
+
−
.
2 12 2 2
2
Observations:
• Since this is a definite integral we do not need to convert back to the original variable x.
• Use the half-angle formulas
cos2 (x) =
to evaluate
R
sin2 (x) dx or
R
1
(1 + cos(2x))
2
and
sin2 (x) =
1
(1 − cos(2x))
2
cos2 (x) dx.
See solution video
Example 4. Evaluate
Z p
5 + 4x − x2 dx.
Intuition: At first glance this appears to be a difficult integral. However, if we complete the square inside
the square root function we will then be able to apply a trigonometric substitution.
Completing the square yields
−x2 + 4x + 5 = − x2 − 4x − 5
= − x2 − 4x + 4 + 9
= −(x − 2)2 + 9
= 9 − (x − 2)2 .
In order to evaluate
Z p
9 − (x − 2)2 dx we set x − 2 = 3 sin(θ) so that x = 2 + 3 sin(θ) and dx = 3 cos(θ)dθ.
3
Calculus II Resources
Z p
Integration Techniques
Z p
9 − (x − 2)2 dx
Z p
=
9 cos2 (θ)3 cos(θ) dθ
Z
= 9 cos2 (θ) dθ
Z
9
=
1 + cos(2θ) dθ
2
9
1
=
θ + sin(2θ) + C
2
2
9
= [θ + sin(θ) cos(θ)] + C
2"
#
p
9
x−2
x − 2 9 − (x − 2)2
=
arcsin
+ C.
+
2
3
3
3
5 + 4x − x2 dx =
Observation: Before converting back to the original variable x, it is useful to rewrite sin(2θ) with the
double angle formula
sin(2θ) = 2 sin(θ) cos(θ).
See solution video
Summary:
• We apply a Trig Substitution to integrals involving the expressions a2 − x2 , a2 + x2 or x2 − a2 .
• A Trig Substitution converts the original integral into an equivalent trigonometric integral.
• When applying a Trig Substitution to an indefinite integral the final step is to return to the original variable. This can be accomplished by sketching the right triangle corresponding to the Trig
Substitution.
• When applying a Trig Substitution to a definite integral you must change the limits of integration.
• Before applying a Trig Substitution you may need to algebraically simplify the integrand or apply a
different technique of integration.
See Trig Substitutions overview video
Practice Problems: Evaluate the following Integrals
Z √
(1)
Z
(2)
1
2
4 − x2
dx
x2
√
x2 − 1
dx
x
Z
(3)
√
Z
(4)
x
x2
p
1
dx
− 6x + 13
1 − x4 dx Hint: Start with an appropriate substitution.
See Solutions
4