#  Trig Sub ```Trigonometric Substitution
Math 122
Three Types of Substitutions
We use trigonometric substitution in cases where applying trigonometric identities is useful.
In particular, trigonometric substitution is great for getting rid of pesky radicals. For example,
if we have √x + 1 in our integrand (and u-sub doesn't work) we can let x = tan θ. Then we
get
2
√x
2
+ 1 = √tan
In this case, we used the identity 1 + tan
2
2
2
θ + 1 = √sec θ = sec θ
θ = sec
2
θ
.
Here are the three types of trigonometric substitutions:
SUBSTITUTION
IDENTITY
x = a sin θ
cos
x = a tan θ
1 + tan
x = a sec θ
1 + tan
2
USE WHEN
θ + sin
2
2
2
√ a2 − x2
θ = 1
θ = sec
θ = sec
2
2
θ
√ a2 + x2
θ
√ x2 − a2
Examples
HW #5
x
2
∫
dx
√1 − 2x2
First we consider whether u-substitution would work here...it doesn't - to see that, let
u = 1 − 2x , and see what happens. To get rid of the radical, we consider the three types of
substitutions and when they apply. In this case, it doesn't quite match any of the three forms
exactly, but it most closely resembles the rst one. We can rewrite the radical term:
2
√1 − 2x
2
2
= √2(1/2 − x ) = √2√1/2 − x
2
Now, a
2
Let x =
= 1/2 ⟹ a = 1/√2
1
√2
sin θ
.
. Then, for the radical term, we get
√2√1/2 − x
2
2
1
= √2√1/2 − (
= √ 2√
sin θ)
√2
= √ 2√
1
(1 − sin
2
1
θ) = √2
2
2
1
−
2
sin
2
θ
√cos2 θ = cos θ
√2
Okay, now we have the radical term: √1 − 2x
dx
dθ
1
2
⟹ cos
2
1
θ
. We also need to nd dx.
1
=
cos θ ⟹ dx =
cos θ dθ
√2
√2
Now we can substitute everything in:
2
x
(
2
∫
dx = ∫
√1 − 2x
To integrate sin
2
θ
sin
1
sin θ)
√2
1
θ
∫
1
dx =
2√ 2
∫
(
1
1
−
2
2√ 2
θ
dx
√2
2√ 2
, we need to use the double angle formula: sin
2
2
cos θ dθ = ∫
cos θ
2
sin
1
cos 2θ) dx =
2
2
θ =
(
2√ 2
1
1
2
−
θ −
2
1
1
2
cos 2θ
sin 2θ)
4
Now to get everything back in terms of x...
sin 2θ = 2 sin θ cos θ
Since x =
1
√2
sin θ
, we have that sin θ = √2x and θ = arcsin 2x.
From our simpli cation of the radical term in the original integral, we have
cos θ = √1 − 2x .
2
1
(
2√ 2
1
θ −
2
1
4
1
=
1
sin 2θ) =
2√ 2
(
1
(arcsin 2θ) −
2
2√ 2
(
1
(arcsin 2θ) −
2
1
(2 sin θ cos θ))
4
1
2
(√2x)(√1 − 2x ))
2
x
2
∫
1
dx =
√1 − 2x2
1
(arcsin 2θ) −
4√ 2
4
x√1 − 2x
2
+ C
HW #7
2
√ 4 − x2
∫
√2
x
2
dx
In this example we have a radical term of the form √a
Let x = 2 sin θ.
2
− x
2
, where a = 2.
The radical term becomes √4 − (2 sin θ)
2
= √4 − 4 sin θ = 2 cos θ
2
For dx we get dx = 2 cos θdθ.
For the lower bound we have √2 − 2 sin θ, so θ = π/4.
For the upper bound we have 2 = 2 sin θ, so θ = π/2.
The denominator term becomes x = (2 sin θ) = 4 sin
2
2
2
θ
.
Substituting everything in we get
2
∫
x
√2
To integrate cot
2
π/2
√ 4 − x2
dx = ∫
2
4 sin
π/4
we use the trig identity cot
θ
π/2
∫
π/2
2 cos θ
2
2
2 cos θ dθ = ∫
θ
cot
2
θ + 1 = csc
2
θ
.
π/2
cot
2
θ dθ = ∫
π/4
(csc
2
π/4
∣
− cot θ − θ∣
∣
θ dθ
π/4
∣
θ − 1) dθ = − cot θ − θ∣
∣
π/2
π/2
π/4
π
= (0 − π/2) − (−1 − π/4) = 1 −
4
π/4
HW #13: Area of an ellipse
In this problem, we are trying to nd the area of the ellipse
x
y
+
9
= 1
4
. To make things a
little easier, let's just nd the area of the top-right quarter and then multiply it by 4.
First, solve for y (for the top half of the ellipse):
x
y
+
9
x
= 1 ⟹ y = 2√ 1 −
4
9
The area of the ellipse will be given by
3
4∫
x
2√ 1 −
0
dx
9
Notice that this can be done using trig substitution. Let x = 3 sin θ.
The radical term becomes √1 −
x
= √1 −
9
9 sin
2
9
θ
= √1 − sin
2
θ = cos θ
The lower bound: 0 = 3 sin θ ⟹ θ = 0
The upper bound: 3 = 3 sin θ ⟹ θ = π/2
The dx term: dx = 3 cos θ dθ
Substituting everything in we get
3
4∫
2√ 1 −
0
To integrate cos
2
θ
π/2
x
dx = 4 ∫
9
θ dθ = 24 ∫
π/2
2
θ dθ = 24 ∫
1
(
0
2
cos
2
θ dθ
0
we need to use the double angle formula cos
cos
0
2(3) cos
0
π/2
24 ∫
π/2
2
2
θ =
2
2
+
1
2
cos 2θ
π/2
1
+
1
cos 2θ) dθ = 12 ∫
(1 + cos 2θ) dθ
0
Integrating we get
π/2
π/2
1
∣
(1 + cos 2θ) dθ = 12 (θ +
sin 2θ)∣
∣
2
0
12 ∫
0
= 6π.
These examples all used the sine substitution; let's go through a few using tangent and secant.
HW #8
1
∫
dx
2
x √16 + x
The radical term in the denominator ts the for √a
substitution.
2
2
+ x
so we'll use the tangent
2
Let x = 4 tan θ.
The radical term becomes √16 + (4 tan θ)
The dx term becomes dx = 4 sec
1
∫
2
θ
2
2
= √16 + 16 tan θ = 4 sec θ
.
1
dx = ∫
4 sec
2
2
sec θ
dθ = ∫
(4 tan θ) (4 sec θ)
2
2
x √16 + x
16 tan
2
dθ
θ
We can simplify a bit here by writing these trig functions in terms of sine and cosine:
sec θ
tan
2
= cot
2
cos
θ sec θ =
θ
sin
2
2
θ
1
=
θ cos θ
cos θ
sin
2
= cot θ csc θ
θ
Now we have
sec θ
∫
16 tan
2
dθ =
θ
1
∫
cot θ csc θ dθ =
16
1
(− csc θ) + C
16
Finally, we rewrite our answer in terms of x:
Our original substitution was x = 4 tan θ. We can rewrite this as tan θ =
x
4
.
Using basic trig, we draw a triangle with [opposite side] = x and [adjacent side] = 4.
From the pythagorean theorem, we get [hypotenuse] = √16 + x .
csc θ = [hypotenuse]/[opposite side] = √16 + x /x.
2
2
In conclusion,
1
∫
√16 + x2
1
dx =
2
2
x √16 + x
(− csc θ) + C = −
16
+ C
16x
HW #9
2
∫
1
√ x2 − 1
x
The radical term in the numerator is of the form √x
substitution.
dx
2
− a
2
, so we'll use the secant
Let x = sec θ
Radical term: √x − 1 = √sec θ − 1 = √tan
dx term: dx = sec θ tan θ dθ.
Lower bound: 1 = sec θ ⟹ θ = 0.
Upper bound: 2 = sec θ ⟹ θ = π/3.
2
2
2
θ = tan θ
.
Substituting everything in we have
2
∫
1
π/3
√ x2 − 1
dx = ∫
x
sec θ tan θ dθ = ∫
sec θ
0
Apply the identity 1 + tan
2
θ = sec
2
π/3
tan θ
θ
0
:
π/3
∫
π/3
tan
0
tan
2
θ dθ = ∫
(sec
2
θ − 1) dθ
0
Now we can integrate directly:
π/3
∫
(sec
0
2
∣
θ − 1) dθ = tan θ − θ∣
∣
π/3
= √3 − π/3
0
2
θ dθ
```